Measure, Integration and Function Spaces

OJ, {t E E :
=

" La C k

Ak1

k=l

then

k


~ akJ1-(Ak n E)

so 'P is J1--integrable over E if and only if Il( Ak n E) < 00 for ak i O. One of the iuteresting properties of the Lebesgue integral is that it is an absolute integral for measurable functions.

Proposition 11 Ltt J1--integrable. In this case,

f :S

lIs fdJ1-1

S

--t

R* be measurable. Then

f

is Il-inlegrable

the last inequality follows.

Ifi

f+

+ f- ,

is

Is If IdJ1-.

Proof: Tbe first part is immediate from the inequalities f"'", fProposition 6 and Corollary 8. Since

and

{=}

s If'

CHAPTER 3. INTEGRATION

86

Corollary 12 Let f, 9 : S -+ R* be measurable with If I ~ 9 Jl-a.e. integrable, then f is Jl-integrable. Proof: Since

Is If I dJl

~

Is gdJl

If 9 is Jl-

[Exercise 12], the result follows from Proposition

11. In particular, if f is Jl-integrable over EEL, then f is Jl-integrable over FE L when FeE. The integral is a linear functional over the class of integrable functions.

Theorem 13 If f, 9 are Jl-integrable, then af + bg is p.-integrable for any a, bE R and Is(af + bg)dp. a Is fdp. + b 1s9dp.. Proof: Since laf + bgl ~ lallfl + Ibllgl, the first statement follows from Proposition 11 and Corollaries 8 and 12. It is easily checked that

Is afdp. Leth=f+g·Thenh

h+-h-

J+

a Isfdp.. f-+g+

g-soh++

+g-=J++g++h-

and from Corollary 8

which gives

Concerning the "size" or "growth" of an integrable function, we have

Proposition 14 Let f : S

-+

R* be p.-integrable. Then

(i) For every a > 0, Ea = {t : If( t)1

~

a} has finite p.-measure,

(ii) f is finite Jl-a. e. (iii) {t: f(t)

=/: O}

has (j-finite p.-measure.

Proof: (i): 00 > Is If I dp. ~ aJl(Ea ). (ii): If Z = {t: If(t)1 = oo}, then for every a

Is If Idp. so I1(Z) = O.

~

k

If I dp.

>0

~

ap.(Z)

3.2. THE LEBESGUE INTEGRAL

87

(iii) follows from (i) since 00

=J O}

{t : f(t)

U{t : If(t)1 ~ Ijk}.

=

k=)

We next establish the other important convergence theorem for the Lebesgue integral, the Dominated Convergence Theorem (DCT). This result removes the very restrictive monotonicity requirement in the MCT. For this we use an important result due to Fatou.

Theorem 15 (Fatou) Let Uk} be non-negative and measurable. Then

Proof: Set hk = infUj : j

~

k}. Then hk llimfk so by the MCT

Concerning the hypothesis and conclusion, see Exercises 14 and 15.

Theorem 16 (DCT) Let fk' f, 9 be measurable with 9 f.L-integrable and such that Ifkl 9 f.L- a.e. If fk -+ f f.L- a.e., then Uk} and fare f.L-integrable with

:s

is

fdf.L = lim

is

ikdf.L.

Moreover, limJs Ifk - fl df.L = O.

:s

:s

Proof: Since Ifkl 9 f.L-a.e. and If I 9 f.L-a.e., these functions are f.L-integrable. Now 9 - fk ~ 0 f.L-a.e. so by Fatou's Theorem

Hence,

Similarly, fk

is + 9 ~ 0 f.L-a.e.

so

fdf.L

~

lim

is

fkdf.L.

88

CHAPTER 3. INTEGRATION

and Thus, limIsfkdll = Isfdll. Since Ifk fl:S; 2g Il-a.e., the last statement follows immediately from the first. We cannot expect the conclusion of the DCT to hold without some condition on the {fd. For example, take f. kC[O,i/.] and Lebesgue measure on [0,1). On the other hand, the domination condition in the DCT is sufficient but is not necessary. C[.-i/2,Hl/2](l)/t. Then f. -+ 0 and For example, take f.(t)

1

00

f.dm

Cn({k + 1/2)/(k - 1/2))

-+

0

[this computation is justified in the next section where it is shown that the Lebesgue integral generalizes the Riemann integral], but there is no Lebesgue integrable function 9 dominating the sequence {Jd [if 9 ~ fl<, then g(t) ~ l/t for t > 1J. In Theorem 9 we showed that the integral of a non-negative measurable function defined a measure. Exercise 26 extends this results to the integral of an arbitrary function having a p-integraL We next consider another important property of the integral as a set function.

Theorem 17 Let f be p-integrable. Then ~

Proof: First, suppose f

o.

Set f.

lim IE fdp

I'(E)~O

f 1\ k. Then 0

If E > 0 is given, choose k such that IsU p(E) < b, then

If f is p-integrable, then lIE fdlll :s; IE the first part.

f.)dp

IfIdll

= O. :s; f. i f so by the MCT,

< 1'/2 and 0 < b < £/2k. If

so the general result follows from

Finally, we have

Proposition 18 Suppose that f has a p-integral over S. If Ie f dp = 0 for every EEL, then f 0 p-a.e. [so f is p-integrable with Is fdll = 0 by Exercise 12}. Proof: If E 10. Similarly, f

= {t: f(t)

~

O}, then IE fdp 0 so f {t: f(t) < O}.

0 p-a.e. in F

= 0 p-a.e.

in E by Proposition

89

3.2. THE LEBESGUE INTEGRAL

Remark 19 We have chosen to basically follow Lebesgue's original construction of the integraL We began with a measure and then constructed the integral from the measure. It is possible to follow the reverse order; this approach to the integral is due to Daniell. In the Daniell approach one begins with an "elementary integral" /, a linear functional defined on a vector space of functions, Z, defined on some subset S, which is positive in the sense that /(1) 2': 0 whenever f 2': 0, fEZ, and which satisfies a mild sequential continuity condition [for example, the Riemann integral on the class of continuous functions]. The functional/is then extended to a positive linear functional J defined on a vector space X of functions on S which contains Z. There is a natural measure associated with the extension J ; 2..: {E : CE E X} is a O"-algebra of subsets of Sand I1(E) J(CE) is a measure on 2..:. Under appropriate assumptions, J(J) = Is fdl1. For treatments of the Daniell integral, see [Roy] or [Ta]. Appendix I: We give a proof of the statement in Remark 5. First, suppose that f is measurable and the range of f is contained in [£, L). Let E > 0 and £ = Yo < YI ••• < y" = L be a partition of [£, LJ with max{£i+l-£,:i=O,I, ... ,n

l}
Set E,={t:£'_l:::;f(t)<£.},i=l, ... ,n,

and

n

n

",p

= L£,CE" V'

;=1

i=1

Then V' :::;

L £,-1 CE ,.

f :::; ",p and n

V')dl1 :::; L(£i - £'-I)I1(E.)

< EI1(S)

i=l

so (1) (in Remark 5) holds. Suppose (1) holds. For each k there exist simple functions V'k and ",pk such that V'k :::; f :::; ",pk and IS(",pk - V'k)dl1 < 11k. Set V' = SUPV'k,,,,p inf",pk. Then V' and ",p are measurable with V' :::; ",p and

k(",p - V')dl1 :::; k(",pk - V'k)dl1 < 11k for each k so Is(",p - V' )dl1 0 and ",p l1-a.e. and f must be measurable.

=

V' l1-a.e. by Proposition 10. Hence",p

=f

V'

Appendix II: Countable Additivity of Lebesgue measure on R". We show, using the MCT, that Lebesgue measure m" on the semi-ring S" IS countably additive (Example 2.2.10). The proof is by induction on n. For n = 1, this follows from Example 2.2.9. Assume the result is true for n. Let

90

CHAPTER 3. INTEGRATION

with Bi E 8 n +1 pairwise disjoint and with union B x E R". Then

A x [a, b) E 8 n + 1 , A E 8 n . Fix

k

L:CA,(X)C[a.,b,)

i CA(X)C[.,b)

i=l

as k

-+ 00

so by the MeT for the measure m, we obtain k

L: CA. (x)(b; -

at)

i CA(x)(b a).

i=l

We now apply the MeT to the measure m n , which is countably additive by the induction hypothesis, to obtain k

k

L:m,,(A,)(b, i=l

at) =

L: m +1(B.) i m,,(A)(b n

a) = mn+l(B).

1=1

Hence, m,,+1 is countably additive. Note that for the proof above we only required the MeT for simple functions. This can be obtained by proving Propositions 1 and 3 and then using the method of proof in Theorem 7 (MeT). As noted earlier in Example 2.2.10 a geometric proof of this result which does not use integration theory can be given, but the proof is surprisingly difficult. Appendix III: As promised following Theorem 2.2.1.5 we give an example of a realvalued, finitely additive set function defined on a u-algebra which is not bounded. For our construction we first require a lemma which utilizes integration with respect to a finitely additive set function as described in Remark 2. Lemma 20 Let A, B be algebras of subsets of S with A c B and let a : A -+ R be finitely additive. If B E B\A and b E R, there exists f3 : B -+ R, finitely additive, such that f3 is an extension of a with f3(B) = b. Proof: Let 8(A) [8(B)J be the vector space of all A-simple [B-simple] functions. Then a induces a linear functional 8(A) -+ R via integration with respect to a, i.e., aU) = fsfda [Remark 2J. The linear functional a has a linear extension, jJ, to 8(B) such that /3(CB ) b. Then f3(E) = !9(CE ), E E B, defines the desired finitely additive extension of a.

a:

We now present our example ([Gil). Example 21 Let {Edk:o be a pairwise disjoint sequence of intervals such that

Ek

= R.

Let Ak be the algebra generated by {Eo, E 1 , ••. , E k } so Ao C Al C A2 C ...

eM,

3.2. THE LEBESGUE INTEGRAL

91

the Lebesgue measurable subsets of R. Set ao 0 on .40; let al be a finitely additive extension of ao t.o At such that. al(Et ) = 1 [Lemma 20], and by induction there exists a sequence {ak} of finitely additive set functions defined on {Ad such that

U

U

ak+l extends ak and ak(Ek ) = k. Now A Ak is an algebra and a ak is k=O k=O finitely additive on A. By Lemma 20, there is a real-valued finitely additive extension of a, p, to M, and since peEk) k for every k, p is unbounded. Exercise 1. If p is counting measure on S, show 1 : S - t R is p-integrable if and only if A = {t : I(t) of O} is countable and I: II(t)1 < 00. In this case, show tEA

Isidp

I:I(t).

tEA

00

Exercise 2. Let

{Jd be p-integrable and I: Is IIkl dp <

00.

Show the series

k=l

converges pointwise ( absolutely) p-a.e. to a p-integrable function

Exercise 3. Let Ik,

Show IE Ikdp

-t

1 be non-negative,

IE Idp for every E E

p-integrable with Ik

-t

I:!k

1 with

1 p-a.e.

Suppose

[Hint: Apply Fatou's Theorem to E and

S\E.] Exercise 4. Let

1 be ,i-integrable.

for every 0 < peE) <

00.

Show

Suppose there exists k

> 0 such that

111 ::.; k p-a.e.

Exercise 5. Let p be finite. Assume {Jd are p-integrable and Ik - t 1 uniformly on S. Show 1 is p-integrable and Is 11k - 11 dp - t O. Can finiteness be dropped? Exercise 6. Let 1, 9 be non-negative, measurable. Assume 1 is p-integrable and set veE) IEIdp for E E Show 9 is v-integrable if and only if Ig is p-integrable and in this case IE gdv IE 19 dp for E E [Hint: First consider simple functions

g.] Exercise 7. Let 1 : [a, b]-t R be m-integrable. If I: I(t)tkdm(t) k 0,1,2, ... , show 1 = 0 m-a.e.

0 for

CHAPTER 3. INTEGRATION

92

Exercise 8. Let f : R -+ R be m-integrable. For a E R set fa(t) = f(t + a). Show fa is m-integrable and fR fdm fR fadm. [See Exercise 2.5.2.] What about fR f(at)dm(t)? [Exercise 2.5.4.] Exercise 9. (Chebychev Inequality). Let p{t: If(t)1 2: e} :s; fs IfI dple. Exercise 10. Let

f be p-integrable and e > O. Show

f be p-integrable. Show limp{t:lf(t)l2:e}

c-+oo

0

and

{

l{t:IJ(tJr?:c}

IfI dp

Exercise 11. Let {Ik}, 9 be p-integrable with

O.

Ilkl :s; 9 p,-a.e.

Show

uniformly for kEN. Exercise 12. Show that if p,( E) over E with fE fdp, = 0.

0, then any measurable function

f is /l-integrable

Exercise 13. Show the analogues of the MCT and DCT do not hold for the Riemann integral. Exercise 14. Show that strict inequality can occur in Fatou's Theorem. [Consider fk = C(O,2) for k odd, fk = C p ,3) for k even.] Exercise 15. Show the non-negativity assumption cannot be dropped in Faton's Theorem. Exercise 16 (Bounded Convergence Theorem; BCT). Let p be a finite measure and fk, f be measurable. Assnme 3M > such that 11;,1 :s; M /l-a.e. If fk -+ f /l-a.e., show fs fdp = limfs fkdp,.

°

k

Exercise 17. Let f : S -+ R* be non-negative and p, (i-finite. Show that measurable if and only if f II 9 is /l-integrable for every Ii-integrable g.

f is

93

3.2. THE LEBESGUE INTEGRAL

L: be pairwise disjoint and E = UE i . Let f be measurable 1=1 and It-integrable over each E i • Show that f is It-integrable over E if and only if Exercise 18. Let {Eo} C

filfl dp < i:::]

00.

E,

Show this last condition cannot be replaced by

li~1 fE, fdPI < 00.

Exercise 19. Suppose f : R R is uniformly continuous on Rand m-integrable over R. Show f vanishes at 00 and is bounded. Can uniform continuity be replaced by continuity? Exercise 20. If f is bounded and measurable and 9 is p-integrable, show fg is p-integrable. Exercise 21. Let p be finite and f measurable. If fg is p-integrable for every Itintegrable g, show there exists M ?:: 0 such that If I :S M p-a.e. [Such functions are called p-essentially bounded.] Exercise 22. Let

f

be non-negative and measurable and set

Ek If f is p-integrable, show

{t:k:Sf(t)
00.

Show

+ l)p(Ek) < 00

k=O

implies that

f

is p-integrable.

Exercise 23. If f is It-integrable and F. = {t: If(t)1

?:: k}, show limkp(F.)

= O.

Exercise 24. Show that E C Rn has Lebesgue measure 0 if and only if there exists a sequence of m-integrable functions U.} such that

f

'=1

~Rn If.ldm < 00

and

IA(x)1 for every x E E.

00,

CHAPTER 3. INTEGRATION

94

Exercise 25. Show the measurability assumption in Proposition 11 and Corollary 12 is important. Exercise 26. Let I : S -> R' have a Jl-integral (possibly infinite). Show v{E) = IE IdJl defines a signed measure on E. Give a Hahn Decomposition for v. Show

i rdJl and Ivl (E)

i

III dJl.

Exercise 27. Let Ik be non-negative, Jl-integrable and suppose Show I 0 Jl-a.e. if and only if HmIs IkdJl = O. Exercise 28. If Jl is a regular measure and that v = IldJl is a regular measure.

I

!kef) 1 I(t) for t

E

S.

is non-negative and Jl-integrable, show

Exercise 29. Let Ik be non-negative and measurable. If Ik :s: I, show

/k

->

I pointwise and

Exercise 30. Let I : [0,1] -> R be Lebesgue integrable. Show t -> t k I(t) is Lebesgue t k I{t)dt -> O. integrable for each kEN and

Id

Exercise 31. Let v be a signed measure on E with v = v+ - v- its Jordan decomposition. 1£ I : S -> R' is E-measurable, say that I is v-integrable if and only if I is both v+ and v- integrable and define

/Idv Show

I

= /Idv+

- /Idv-.

is v-integrable if and only if I is lvi-integrable and in this case

1/ Idvl :s: / Show Ivl(E)

I/ldlvl·

= sup {IIEldvl : 1/1:s: I}.

Exercise 32. Let Jl be a finite measure and ft, non-negative and Jl-integrable with Ik -> 0 Jl-a.e. Show Is IkdJl -> 0 if and only if for every c > 0 there exists 8 > 0 such that Jl (E) < 8 implies IE IkdJl < c for all k.

3.2. THE LEBESGUE INTEGRAL

Exercise 33.

95

Show p. is a-finite if and only if 3 a p.-integrable function E S.

f with

f (t) > 0 for all t

Exercise 34. If f (t) > 0 for all tEE, p. (E) > 0 and IEfdp. > O.

f is p.-integrable over E, show

Exercise 35. Let f : Rn -> R" be Lebesgue integrable. If IK f dm compact (open) K, show f = 0 m-a.e. Exercise 36. Find limI~

0 for every

(i!:;.dt.

Exercise 37. Let Ud be measurable and !k -> f p.-a.e. Let 9k, 9 be p.-integrable, gk -> 9 p.-a.e. and Is 9k dP. -> Is 9dP.. If Ifk I S 9k p.-a.e., show f is p.-integrable and Is fkdp. -> Is fdp.. Exercise 38. Let f : [a, b] -> R" be Lebesgue integrable and a S x S b. Show f = 0 m-a.e.

I: fdm = 0 for every

CHAPTER 3. INTEGRATION

96

3.3

The Riemann and Lebesgue Integrals

Let f : [a, b] -> R be bounded. If 7r {a = Xo < XI < ... < Xn b} is a partition of [a,b], set 0; = [Xi-I, mi = inf{J(t) : t E Oil, M; = sup{J(t) : tEo;} and J.l (7r) = max {Xi - xi-d. The upper (lower) sum of f with respect to 7r is n

=L

U(j,1r)

Mi(Xi

Xi-I)

;=1 n

E

[(L(j,1r)

xi-t)L and the upper (lower) integral of f is

mi(Xi

;=1

JJ

-b

[f~f

sup{(L(j,1r):

t,1

1r

inf{U(j,7r): 7r is a partition of [a,b]} a partition of [a,b]}]. The function f is Riemann integrable

[~ver [a, b]] if =f~f, and the Riemann integral of f is defined to be the common value; in order to distinguish the Riemann integral from the Lebesgue integral, we denote the Riemann integral by R f. We now show that any Riemann integrable function is Lebesgue integrable and the two integrals agree in this case. Our proof also gives a necessary and sufficient condition for a function to be Riemann integrable.

f:

Theorem 1 Let f : [a, b]

->

R be bounded.

(i) Iff is Riemann integrable over [a, b], then f is Lebesgue integrable over [a, b] and Rf: f f: fdm.

(ii) f is Riemann integrable over [a, b] if and only if f is continuous m-a. e. in [a, b]. Proof: Choose a sequence of partitions, {IT',,}, such that IT'1 C 1r2 C .. " J.l( 1rk) - t 0 and limL(f,lT'k) = =-
" Pj]. U

satisfy the last three conditions and then setting 1rk

{a

Xo

Xi-l ::;

<

XI

< ... <

Xn

= b},

let

mi

= inf{J(t)

t ::; xt} and define simple functions

:

i k and

Xi-I::;

t ::; Xi}, Mi n

Uk

by i"

E i=1

f

MiC[x,_I,Xi) on [a,b) and i,,(b)

£=1

U(j, IT'k). Since

IT'HI

:J 7rk, we have

u,,(b)

= f(b)

If 7rk is given by

j=l

so f:i"dm

sup{J(t):

miC[:r._l,x.),

u" =

L(j,lT'k), f:ukdm =

3.3. THE RIEMANN AND LEBESGUE INTEGRALS Set f(x)

= limfk(X), u(x)

97

limuk(X) and note u(x) ::::: l(x). By the MCT,

1 b

1

J/.

b-';'

udm

= lim

Uk dm

=

(3.2)

For the proof of (i), assume that 1 is Riemann integrable. From (2) we obtain R 1 udm so (u - f)dm = 0 and since u ::::: f, u = l m-a.e. so from (1), u l = 1 m-a.e. Hence, 1 is Lebesgue integrable and f: ldm Rf: 1.

f: Um

f:

f:

f:

For the proof of (ii), let C =

U

'Irk

so C is countable and m(C) = O. If x

if:

C,

'\;=1

then 1 is continuous at x if and only if u(x) l(x) = lim(uk(x) lk(X» = O. Hence, if 1 is continuous m-a.e., then from (1), u l = 1 m-a.e. and from (2) Um udm = so 1 is Riemann integrable. If, conversely, 1 is Riemann integrable, then from (2) udm = Um so u l m-a.e. and 1 is continuous m-a.e. This theorem now allows us to compute the Lebesgue integral for a large class of functions, and we will freely use properties of the Riemann integral to calculate Lebesgue integrals. We can use Theorem 1 to establish the analogue of the BCT for the Riemann integral.

f:

f:

1,.1

I:

I:

Corollary 2 (Arzela) Let 1,\;, 1 : [a, b] -> R be Riemann integrable with h, -> 1 pointwise on [a,bJ. 11 there exists M > 0 such that 11k(t)1 M lor all k, t E [a,b], then

s

lim R

l fx = l R

j.

Proof: Theorem 1 and the BCT. Note that we must assume the Riemann integrability of the limit function 1 in Corollary 2 [Exercise 1J. Our proof of Arzela's Theorem, of course, depends on measure theory and the Lebesgue integral; for an elementary proof not using such machinery see [Lew]. Improper Riemann Integrals: If 1 : [a, (0) -> R is Riemann integrable over [a, b] for each b > a, the improper Riemann integral of lover [a, (0) is defined to be lim Rib 1 = R [00 1, b-+oo

a

J«

provided the limit exists and is finite. [This integral is also called the Cauchy-Riemann integral.] If R faoo 111 exists, it follows from Theorem 1 and the MCT and DCT that 1 is Lebesgue integrable over [a, (0) and faoo ldm R fa= 1. However, a function can be Cauchy-Riemann integrable over [a, (0) and not be Lebesgue integrable. For example, consider R It' .i~x dx. Integrating by parts, gives

CHA.PTER 3. INTEGRATION

98

and since Icosxl/x 2 :::;1/x 2 , 'RIt si~xdx exists. On the other hand, """sinx, n~11(k+l)"'lsinxl n~1 1 l(k+I)" ,,~I 2 dx = dx > Isin x Idx = ", X k=1 kJr X -k=l(k+l)7l' kJr k=l(k+l)7l'

l

L:

-

L:

L: -:-::---:-:--

so .i~x is not Lebesgue integrable over [1,00). Similar remarks apply to improper Riemann integrals for unbounded functions on bounded intervals.

Defects in the Riemann Integral: I. The first defect is the lack of good convergence theorems. ror example, there exists a uniformly bounded sequence of Riemann integrable functions which converge pointwise to a function which is not Riemann integrable (Exercise 1.3.1). II. Closely related to the lack of strong convergence theorems is the incompleteness of the space of Riemann integrable functions. Let 'R[a, b] be the vector space of Riemann integrable functions on [a, bJ and define a semi-metric d on 'R[a, b] by d(f, 9) If - 91· Convergence in this semi-metric is called convergence in mean and will be studied for the Lebesgue integral in §3.5. We show that d is not complete; in §3.5 we show that the analogous space for the Lebesgue integral is complete. Let H c [0,1] be a 1/2-Cantor set. Let {h} be the open intervals making up [0, IJ\H and set

I:

n

In =

U h,
C Jn •

k=1

Then each
t

l

0

so { 0, (to - E, to + t)\H contains an interval so (to - E, to +t) must contain points t such that f(t) = 1. lIence, limf(t) = 1 for every to E [0,1]. But f =

I:

3.3. THE RIEMANN AND LEBESGUE INTEGRALS

99

intervals making up [0, 1]\H. Define J on (a, b) by setting J(t) "" (t a)2 sin(I/(t-a)) on (a,a) where a < (a + b)/2 is such that /,(a) 0, letting J be the constant J(a) on [a, (a + b)/2] and defining J on [(a + b)/2, bJ by reflection in the line t = (a + b)/2. [A sketch is helpful.] We extend J to [0,1] by setting J(t) = 0 for t E H. Then /' exists everywhere in [O,I]\H and IJ'(t)l:::; 1 for t E He. We claim that f'(x) = 0 for x E H. Let € > 0 and suppose Ix - tl < €. If t E H, then (J(t) J(x»/(t - x) O. If t f. H, then t belongs to some open interval (a, b) making up [O,I]\H. Suppose a is the endpoint nearest x. Then

1(J(t) - J(x»/(t - x)1

= IJ(t)/(t - x)1 :::; IJ(t)/(t - a)1 :::; It

a1

2

/

It - al <

f

Hence, J'(x) O. Thus, J' exists everywhere and is bounded. But, /' is discontinuous on Hand m(H) = 1/2 so /' is not Riemann integrable.

Exercise 1. Show the Riemann integrability of the limit function dropped in Corollary 2. Exercise 2. Let K

c

J cannot be

[0, I] be the Cantor set. Is CK Riemann integrable?

Exercise 3. Let 0 < f < 1 and K. an f-Cantor set. Is CK, Riemann integrable? Exercise 4. Show the following functions are Lebesgue measurable and determine whether they are Lebesgue integrable.

t:::; 1, p > 0,

(a) J(t)

IjtP, 0 <

(b) J(t)

= 0 otherwise, (_1)k/2k for t E [k -1, k) and J(t) = 0 otherwise,

(c) J(t) =

(_I)k /k for t E [k - 1, k) and J(t)

(d) J(t) = l/t for 0 < t < 1, -1/~ for 1 < t < 2 and J(t)

(e) J(t)

IjtP,l:::; t <

00,

p

0 otherwise,

> O.

Exercise 5. Let A be all subsets of [a, bl such that C A is Riemann integrable (Exer. CA. Show p, is countably additive. 2.1.8). Define p, on A by p,(A)

f:

100

CHAPTER 3. INTEGRATION

3.4

Integrals Depending on a Parameter

Let (S, 2::, Ji) be a measure space and I of- 0. If f: SxI ---> R., we write fe,t) [f(s,·)] for the function f(·,t)(s) = f(s,t) [f(s,·)(t) = f(s,t)]. If f(·,t) is Ji-integrable for every tEl, we say that the integral F( t) = fs f( s, t )dJi( s) depends on the parameter tEl. In this section we study properties which the function F inherits from f. First, we consider continuity. Theorem 1 Let I be a metric space and f : S x I

--->

R. Assume

(i) f(" t) is Ji-integrable for every tEl. (ii) f(s,.) is continuous at to E I for each s E S. (iii) There exists a Ji-integrable function 9: S ---> R such that If(s, t)1 s E S, tEl. Then F(t) = fsf(s,t)dJi(s) is continuous at to.

:s: 9(S) for

all

Proof: Let {td be a sequence from I converging to to· Then f( s, td ---> f( s, to) for every s E S by (ii). By (iii) If(-' tk)1 :s: 9 for every k so the OCT implies F(tk) ---> F(to) and F is continuous at to. As an example we consider the Gamma function. Example 2 (Gamma Function) The Gamma function is defined by

for x> O. First, we observe that the integral exists. For 0 < t :s: 1, tx-le- t :s: t x- l and the function t ---> t x- l is integrable over [0,1] for x > 0 (Exer. 3.3.4) so fOl tX-le-tdt is finite. For t > 1, tx-le- t = [tx+le- t ]C 2 and the function t ---> tx+le- t is bounded since lim t x+l e- t = 0 so ftO t x- l e-tdt < 00 (Exer. 3.3.4). Hence, fooo t x- l e-tdt < 00. t~oo

Next, we show that r is continuous for x > O. Let xo > O. If 0 :s: t :s: 1 and xo/2 < x, then t X :s: t xo / 2 and e- t :s: 1 for t ~ O. Therefore, tx-le- t :s: t(x o/2)-l so by Theorem 1, x ---> f~ t x- l e-tdt is continuous at Xo. For t > 1 and xo/2 :s: x :s: 2xo, there is a B such that t x - l e- t :s: Bt- 2 so by Theorem 1

is continuous at Xo. Hence, r is continuous at Xo. Next, we consider differentiability. Theorem 3 (Leibniz) Let I be an interval in Rand f : S x 1---> R. Assume

3.4. INTEGRALS DEPENDING ON A PARAMETER

101

(i) f(-, t) is j.t-integrable for every t E I. (ii) ¥t(s,t) exists for every s E S, t E I. (iii) There exists a j.t-integrable function 9 : S

->

R such that

I~~ (s, t)1 ~ g(s) for s E S, t E I. If F(t) = Is f( s, t)dj.t( s), then F is differentiable on I with F'( t) = Is ¥t(s, t)dj.t(s). Proof: Fix t E I and let tk

t, tk E I, tk # t. Then for each s E S,

->

lim f(s, tk) - f(s, t) = af (s, t) and f(', tk) - f(', t)

tk - t

k

at

tk - t

is j.t-integrable for each k. By the Mean Value Theorem, for each s, k there is a between tk and t such that

Zs,k

f(s,tk) - f(s,t) = af(s Z ) so If(s,t k ) - f(S,t)1 < (s) tk - t at' .,k tk - t - 9 for every s E S by (iii). Hence, the DCT implies lim

k

f f(s,tk)-f(s,t)dj.t(s) = f ~f(s,t)dj.t(s)=F'(t).

15

15

tk - t

ut

As an example of how Theorems 1 and 3 can be used, we evaluate the integral

Iooo e- x2 dx.

Example 4 (Euler) Since 0 ~ e- x2 ~ e- X for x 2 1, e- x2 is Lebesgue integrable over [0,00). For t 2 0, set f(t) = (J~e-X2dx)2 and get) = I~e-t2(x2+1)/(x2 + l)dx. From standard results on Riemann integration, f is differentiable on [0,00) with

Since

I:t (e- t2 (x2

+I) /(x

2 + 1))

1= /_2(te-t2)e-t2x2/ ~ B

for t 2 0, 0 ~ x ~ 1, Theorem 2 implies that 9 is differentiable on [0,00) with

g'(t) Setting u = tx for t

= _2te- t2 !al e- t2x2 dx.

> 0 in this last integral gives g'(t)

= -2e- t

2

lot eo

u2

duo

CHAPTER 3. INTEGRATION

102

Hence, f'(t) + g'(t) 0 for t > 0 and f(t) + get) = c for t > O. Now f is continuous on [0,(0) and since ie-t2(x2+1)/(x2 + 1)1:::; 1/(x 2 + 1) for t 2: 0, 0 :::; x :::; 1, 9 is also continuous on [0,(0) by Theorem 1. Also, by the DCT, lim t-+oo

Thus,

f +9

1 1

0

) e- t " (x +1 /(x 2 + l)dx

con [0, (0) and

f(O)

+ g(O)

I~ I~x,dx lim (f(t)

t .....

Exercise 1. For x > 0 show r( x rex + 1) xl. Show 1'(1/2) = fo. Exercise 2. Show F( x) Exercise 3. Example 4.

O.

Show

10

00

=

+ 1)

It' x~i~:2 dt x 2n e- x2 dx

=

7r/4

Uo= e-

+ get))

X

'dx)2

xr( x). If x is a positive intcgcr show

is continuous for x E R.

(2n)!fo/(22nnl2).

Hint: For n

o this

is

Exercise 4. Show Ioooe-tx'dx = V(7r/t)/2 for t > O. Exercise 5. Show the function F(x) =

10

00

e- xt /(1

+ t)dt

is differentiable for x> O.

Exercise 6. Show the Gamma function is differentiable. Exercise 7. Let F(t) 10= e-txsi~xdx for t > O. Show F(t) = i-arctan t. Hint: FI(t) = -1/(1 + t 2 ) so F( t) c - arctan t. Evaluate c by considering {F( n

n.

Exercise 8. Evaluate F(x) = I;(t" -1)fln tdt for x > O. Hint: Find FI(x) and note

F(x)-+Oasx-+O.

8.5. CONVERGENCE IN MEAN

3.5

103

Convergence in Mean

In this section we consider the approximation of integrable funcLions wiLh respect to a natural semi-metric induced by the integral. Let p.) be a measure space. \Ve denote by LI t.he space of all functions I : S ---> R which are It-integrable. By :.3.2.13 L 1 (p) is a vector space (under the usnal operations of pointwise addition and scalar multiplication). We defiue a semi-metric d1 on V (fl) by d1 (log) f, II gl dp,; note d l g) = 0 if and only if I g fl-a.e. For cOllvenience we set Ilflll = Is Ifl then d l (f,g) = III - gill' II III called the Llnorm of I; we consider more general such norms in §6.1. Convergence in this semi metric is called p,-m.ean convergence or convf'1:qcnce in p.-mean. We often wrik lk -> .f IHlIean wh('n Illk 1111 -+ O. If 1 is an int.erval in R", we deuote by [,1 (I) the space o[ real, valued integrable functions on [when I is equipped wit II the Lebesgue measme. We COlllpare mean convergence with other modes of convergence in §:1.7. On(' of Lhe most important properties of V (/1) its completeness with respect to the mdric of mean convergence.

Theorem 1 (Riesz- Fischer) V (,t} is complete. Proof: Let {nd such that,

{Id be Cauchy in [}(p). Then 111'1<+1 Inklll < 1j'2k. Since

there exists an increasing sequence

- Ink Idp

00.

the series

converges It'a.e. to a real-valued fundion which belongs to V (p) (Exer. Therefore, the series

.ft + "IJlnk+l

fnk)

limfnk

3.2.2).

1

k=l

converges p-a.e. But } also (,onvt':'rges in mean to I since given any E > 0, Is lInk -rlfl E [or large k, j, and by Fatou's Lemma, letting j --+ 00 gives Is Ifnk ,- 11 dp, ::; E for large k. This mean;.; that I E Ll(/l) and In. -+ 1 p-mean. Hence, !k . . ."' .f p,-mean.

I

CHAPTER 3. INTEGRATION

104

The analogue of Theorem 1 for the Riemann integral is false (§3.3), and its failure is perhaps the most important reason for the overwhelming use of the Lebesgue integral. We now consider some dense subspaces of L 1 (fl): Dense Subsets of Ll(fl): Theorem 2 The vector space of'L-simple fl-integrable functions, S('L), is dense in Ll(fl) (with respect to d 1). Moreover, given fELl (fl) there exists a sequence of simple functions {'I'd in Ll (fl) such that rpk -+ f pointwise and IIrpk fill -+ 0,. if f 2: 0, the {'I'd can be chosen such that rpk if·

Proof: If f E Ll(fl), pick a sequence of 'L-simple functions {rpn} such that rpn -+ f pointwise with 1'1',,1 ::; If I (3.1.1.3). The OCT implies d1 (rpn, f) -+ 0. The last statement follows from 3.1.1.2. For the next result assume that fl is a premeasure on a semi-ring S of subsets of S and that 'L is the a-algebra of fl"-measurable subsets of S (§2.4). Let fl denote the restriction of fl" to 'L. Theorem 3 The vector space of S -simple fl-integrable functions is dense in Ll (fl). Proof: From Theorem 2 it suffices to show that for each E E 'L with fl(E) < 00 and each I' > there is an S-simple function 'I' such that liCE - rpil l < c Pick

°

{AJ C S pairwise disjoint such that

Aj ::) E and

f

fl(Aj)

< fl(E)

;'=1

Set B

UAj and note fl(B) <

00

and

f

= CB .

CA

j=1

j=1

+ (/2 (2.1.11).

Choose N such that.

'

N

LCA,)dfl=

L

fl(Aj) <1'/2.

j=1

Then

<

Is ICE fl(Aj)

CBI dfl

+ Is I CB -

fl(E)

j~1 C Idfl A,

+ 1'/2 < c

We next consider the approximation of integrable functions by continuous functions. A topological space S is locally compact if every point in S has a neighborhood with compact closure. For example, Rn is locally compact. For locally compact Hausdorff spaces, we have the important lemma of Urysohn. Lemma 4 (Urysohn) Let S be a locally compact Hausdorff space, K C S compact and V open with K C V. Then there exists a continuous function f : S that f(t) 1 for t E K and f(t) for t f/c v.

°

-+

[0, 1] such

3.5. CONVERGENCE IN MEAN

105

See lSi] §28 for the proof in a general topological space. For metric spaces (such as Rn), one can use the function f(t) = dist( t, VC)/( dist(t, K) + dist( t, VC». We need a slight refinement of Urysohn's Lemma. If f : S --> R is a continuous function, the support of f, denoted by spt(f), is the closure of the set

{t E S: f(t)

i=

O}.

The space of continuous functions on S with compact support is denoted by Lemma 5 Let S be locally compact, Hausdorff. If K is compact and V is open with 1 for t E K and K C V, then there exists f E Cc(S) such that f : S --> [0,1], f(t) f( t) = 0 for t E VC. Proof: If S K, trivial so assume x E S\K. For each y E K there is an open neighborhood Ny C V of y with compact closure and an open neighborhood Uy of x such that Ny n Uy 0. Then {Ny ; y E K} is an open cover of K and, therefore, j

has a finite subcover, NJ, .. . , N j with V

::::>

U

= U N.

::::>

K and since

is compact,

i=]

::::> K is compact. By Lemma 4 there is a continuous function f : S --> [0,1] such that f(t) = 1 for t E K and f(t) = 0 for t E uc. Since spt(f) C V, f E Cc(S).

V

Theorem 6 Let S be locally compact Hausdorff and let 11 be a Borel measure on 8(S) such that every Borel set is inner regular. Then Ce(S) is dense in LI(I1)' Proof: By Theorem 2 it suffices to show that CB for B E 8(S) and I1(B) < 00 can be approximated by a function in Cc(S). Let (' > O. Since B and Be are inner regular, there exist KI C B compact with I1(B\Kd < 1')2 and K2 C Be compact with I1(Be\K 2 ) < f/2. By Lemma 4 there is a function f E Ce(S) such that 0 S; f(t) S 1 for t E S, f(t) 1 for t E KI and f(t) = 0 for t E K 2 • Then

is If -

CBI dl1

Note that this theorem is applicable to Lebesgue and Lebesgue-Stieltjes measures (Exer. 2.5.10 and 2.6.5). See also Proposition 2.7.2. Exercise 1. Show Ll is not, in general, closed under pointwise products. [Hint: Consider t a for 0 S; t S; L] Exercise 2. Show the polynomials are dense in L1[a,b]. Generalize to Rn. Exercise 3. Show

VIa, b]

is separable. Generalize to Rn.

Exercise 4. Give an example of a non-regular measure for which Cc(S) is not dense in £1(11). [Hint: Use R with Lebesgue measure and the discrete metric.]

CHAPTER 3. INTEGRATION

106

Exercise 5. What is the completion of R[a, bJ? Exercise 6. Let JL be a finite measure on the o--algebra on I: by d(A,B)

JL(A.6B)

Show d is a complete semi-metric.

=

Define a semi-metric d

is ICA - CBld/t.

3.6.

107

CONVERGENCE IN MEASURE

3.6

Convergence in Measure

\Ve consider another type of convergence with respect to a measure. Let (S, IJ) be a measure space and fk' f : S - t R L:-measurable functions. We say that Uk} converges to f in IJ-measure, fk - t f IJ-measure, if for every a > 0,

lim/t{t: Ifk(t) k

f(t)l2: a}

O.

We compare the various modes of convergence in §3.7. We now develop some of the properties of convergence in IJ-measure.

Proposition 1 Let fk' f, gk, 9 : S

(i) If fk f II-measure and gk for a, bE R. (ii) If fk

0 IJ-measure and gk

(iii) If /t(S) <

00

and fk

(iv) If 11(5) < 00, fk It-measure.

(v) If fk

-t

-t

-t

-t

R be L:-measurable.

9 IJ-measure, then afk

-t

0 IJ-rneasure, then fkgk

f IJ-rneasure, then fkg

f IJ-measure and gk

f IJ-measu.re and fk

-t

+ bgk - t af + bg IJ-rneasure

-t

-t

-t

0 IJ-measure.

fg It-measure.

9 It-measure, then fkgk

9 II-measure, then f

=9

fg

IJ-a.e.

Proof: (i): For a > 0,

IJ{t: l/k(t)

+ gk(t) -

f(t) - g(t)l2: a}

II{t: l/k(t) - f(t)l2: a/2}

+ implies that fk + gk (ii) follows from

-t

(iii): Since It(S) <

f

00,

+9

Il{t: Igk(t) - g(t)l2: a/2}

IJ-measure. That afk

liFIJ{t: Ig(t)l2: k}

--t

af IJ-measure is clear.

0 (2.2.4). Let

E>

O. There exists N

such that II{t: Ig(t)l2: N} < E/2. Now

/l{t: l!k(t)g(t)

f(t)g(I)I2: a}:::: It{t: l!k(t)

f(t)I2: a/N}

so if ko is chosen such that k 2: ko implies IJ{ t : Ifk( t) for k 2: ko that IJ{t: Ifk(t)g(t) - f(t)g(t)l2: a} < E. (iv) follows from (i), (ii), (iii) and the fact that

/kgk - fg

= Uk

- f)(gk - g)

+ f(gk

+ p{t:

Ig(t)l2: N}

f (i)1 2: a / N} < £'./2, we have

g)

+ g(/k

- f).

CHAPTER 3. INTEGRATION

108 (v): Since

{I: f(t)

¥ g(t)} = U {I:

If(l) - g(t)1 ~ 11k},

k=l

it suffices to show that fL{t: If(t) - g(I)1 ~.,.}

=

°

for.,.

> 0.

But

{t: If(t) - g(t)1 ~ oj c {t: Ifk(t) - f(t)1 ~ o"/2} u {t: Ifk(t) - g(t)1 ~ u/2}. Example 2 The finiteness condition in (iii) and (lv) cannot be dropped. Let. S and consider Lebesgue measure. Let fk(t) 11k and get) t for IE R. Then!k in m-measure but for.,. > 0, m{l: 1!k(t)g(t)1 ~.,.} = 00.

=R ---+

0

Concerning convergence in measure and convergence a.e., we have an irnpor1.ant result of F. Riesz.

Theorem 3 (F. Riesz) Let fk such that fn. f fL-(1. e.

f Ii-measure. Then ther'e is

Proof: For each k there exists

nk

such that j ~

nk

(1

.571b.5f:querw: {f"k

)

implies

Set

Ek If t rf:.

UE

k,

= {I:

then Ifn,(t) - f(t)1

Ifn.(t) - f(t)1 ~ 1/2k }.

< 1/2 k for k ~ j. Thus, if

k=j

t rf:.

nU

J~I

then fnk(t) But

---+

Ek

= limEk

E,

k=)

fnk -, f pointwise on S\E.

f(t),

00

k=j

for every j so fL(E)

= 0.

The assertion that there is a subsequence which converges a.e. in Theorem a is important -. the entire sequence may not converge a.e. as the followilLg example shows.

Example 4 Let S = [0,1] and A;; = [(k 1)/2 n, k/2n] for k = 1, ... ,2" and n E N. Consider the sequence CAl, CAl, C A2! ... [see the sketch below]. 2 2 I

1

3.6. CONVERGENCE IN MEASURE

109

This sequence converges to 0 in m-measure since mit : CA.(t) ~ a} :s; m(Ak) 1/2". However, t.his sequence doesn't converge t.o 0 at. any point since it is 1 infinit.ely often at. any point. [See Exercise 4.] We can formulate a Cauchy-type condition for convergence in fl-measure. The sequence U'} is said to be Cauchy in fl-measure if and only if for every a > 0 and f > 0 there exists N such that k, j ~ N implies

We now establish a completeness type result for convergence in fl-measure which is also due to F. Riesz. J~emma

5 If {fk} is Cauchy in fl-measure and has a subsequence {fn.} which converges in fl-measure to a measurable function f, then !k --+ f fl-measure. Proof: Let a,

f

> O. There exists N such that k, j fl{t: Ifk(t) - h(t)1

Choose nk

~

~

~

N implies

a/2} < E/2.

N such that fl{t: Ifn.(t) - f(t)1 ~ a/2} < f/2.

Then for j

~

N,

f(t)1 ~ a}

fl{t: Ih(t)

< fl{t: Ih(t) - fn.(t)1

+

~

a/2}

fl{t: Ifn.(t) - f(t)1 ~ a/2}

<

E.

Theorem 6 (F. Riesz) Let Ud be Cauchy in Il-measure. Then there exists a measurable function f such that fk --; f Il-measure. Proof: For each k there exists

\Ve may assume nk

<

nk+l'

such that i, j

~ nk

implies

Let

1/2 j

so fl(Ek) < 1/2k. If Fk

= O. We claim that Un.}

nk

1/2 k- 1 so if A

IimEk, then Il(A) some k so if i, j

~

converges pointwise on S\A. If t E S\A, then t r/;

k with i

> j,

U

for

J'=k

(3.1)

CHAPTER 3. INTEGRATION

110

and {In. (t)} converges. Now set f(t) limfn.(t) if t E S\A and f(t) 0 for tEA. Then f is measurable since f CS\Alimfn •. Finally, we claim that fn. -+ f fL-measure. Let IJ', t > O. Choose N such that fL(FN) < Ij2 N- 1 < min(lJ', t). If j ;::: N, then

since if t ;f:. FN and j ;::: N, passing to the limit in (1) as i

gives If( t) - fn, (t)1 ~

-+ 00

Ij2 N- 1 • Thus, if j ;::: N, fL{t: If",(t) - f(t)l;::: u} ~ fL(FN) < Lemma 5 now gives the result.

t.

Metric of Convergence in Measure: We now describe a semi-metric which characterizes convergence in measure for finite measures. Assume henceforth that fL is a finite measure. We first require a lemma.

b E R '1+laHI ~ < J<>L L emma 7 I'f J a, - I+lal

+~ 1+lbl'

Proof: Note the function h(t) tj(1 + t) is increasing for t b have the same signs, we may assume that a, b ;::: 0 so

> -1. First, if a and

la + bl j(l+/a + bl) = (a+b)j(Ha+b)

j(l+lal)+lbl j(Hlbl).

~

aj(Ha)+bj(Hb)

/a/

On the other hand, if a and b have different signs, we may assume that Then la + bl ~ la/ implies

la + bl j(1 + la + bl)

~

la/ ;::: Ibl.

lal j(1 + lal) ~ lal j(1 + laD + Ibl j(1 + Ibl)·

Definition 8 Let LO(fL) be the vector space of all real valued measurable functions on S. For f, 9 E LO(fL), set

d(J,g)

( If - gl

= 1s 1 + If _ g/d fL

[note the integral exists since fL is finite and the integrand is bounded]. By Lemma 7, d defines a semi-metric on LO(fL) which is translation invariant in the sense that d(J, g) = d(J + h, 9 + h) for any h E LO(fL) and is such that d(J,O) = 0 if and only if f = 0 {L-a.e. We show that convergence in the semi-metric d is exactly convergence in fLmeasure.

(i) fk

-+

f in fL-measure

{;>

dCA, J)

-+

O.

3.6. CONVERGENCE IN MEASURE (ii) Uk} i!> Cauchy in p,-mea!>ure

¢:>

III

Uk} is Cauchy with respect to d.

Proof: (i): Let c> 0 and set Ek

l(t)1 2: f}. Then

{t: I/k(t)

(3.2) ~

Thus, if fk --t Since h(t)

I

p,(Ek) + tp,(S\E k)

~

p-measure, (2) implies that d(Jk, J) + t) is increasing for t > -1,

p,(Ek) + cp,(S). --t

O.

t/(l

d(f f) > ( I"

-

Ifk

II

iE. 1 + Ilk II

d > _c_ P (E) p, - 1 + c k·

Thus, if d(Jk, J) --t 0, (3) implies that Ik --t f p,-measure. (ii) follows from the inequalities (2) and (3) with f replaced by

(3.3)

h

Corollary 10 d is a complete semi-metric on I.,o(p,). Proof: Theorems 6 and 9.

If I is a bounded interval in Rn, we denote by the space of all Lebesgue measurable functions on I, and we assume that the semi-metric on LO(!) is the semimetric of convergence in Lebesgue measure. There is a semi-metric which characterizes convergence in measure for infinite measures, but it is much more complicated than the semi-metric dj see [DS] IIL2 for a description. Exercise 1. 1£ p, is counting measure on S, show convergence in p,-measure is exactly uniform convergence on S. Exercise 2. measure. Exercise 3.

Show that if fk

If!k

--t

f

--t

I

p,-measure and

p,-measure, show

Ud

I

9 p-a.e., then fk

--t

9 J1-

is Cauchy in p,-measure.

Exercise 4. Find a subsequence in Example 4 which converges m-a.e. to O. Exercise 5. Can the condition "p,(S) is bounded"?

<

(Xl"

in Proposition 1 (iii) be replaced by "g

Exercise 6. Let I : S --t R be measurable. Show I is p,-integrable if and only if there exists a sequence of p,-integrable simple functions {if'd such that (i) if'.

--t

I

p-measure and

(ii) I.imJslif'k •J

if'.il dp, = O.

CHAPTER 3. INTEGRATION

112

3.7

Comparison of Modes of Convergence

In this section we pause to compare the various modes of convergence which have been introduced. Let (S, L:;, fl) be a measure space. We have considered the following types of convergence for sequences of measurable functions defined on S. unif: a. unif: a.e.: mean: meas:

uniform convergence on S. almost uniform convergence with respect to fl (Definition 3.1.16). almost everywhere convergence with respect to fl (3.1.13). convergence in fl-mean (§3.5). convergence in fl-measure (§3.6).

We have the obvious implication that unif =? a. unif and from Exercise 3.1.13 we have a. unif =? a.e. It is also clear that a. unif =? meas 1). There is one other general implication which we now establish.

Proposition 1

Illk

-->

I

fl-mean, then

Proof: Let 17 > 0 and Ek

r Ilk -

Js

Ik

-->

I

{t: I/dt) - l(t)1

II dfl ~

fl-measure. ~

17}. Then

r Ilk - II dfl ~ 17fl(Ek) so fl(J~k) -> O.

JR.

We now give examples to show that these are the only possible general implications which are valid. [Recall from Egoroff's Theorem a.e. =? a. unif for finite measmes.] 2. a.e. f;- a. unif: Take

!k

C[k,oo)

3. a. unif f;- unif: Take Ik(t)

in R with Lebesgue measure.

tk, 0 'S t 'S 1, and Lebesgue measure.

4. meas f;- a.e.: Example 3.6.4. 5. meas f;- a. unif: Example 3.6.4.

6. meas f;- mean: Take

Ik

kC[O,l/klon [0,1] with Lebesgue measure.

7. a.e. f;- mean: Same as 6. 8. mean f;- a.e.: Example 3.6.4. 9. a.e. f;- meas: Same as 2.

10. a. unif f;- mean: Same as 6. 11. unif

mean: Take Ik

qo,kl/k on R with Lebesgue measure.

3.7. COMPARISON OF MODES OF CONVERGENCE 12. mean

p

113

a. unif: Example 3.6.4.

We can summarize the relationships above by means of the chart:

unif

~.a.

unif

/ meas

a.e.

If 11 is a finite measure, then by Egoroff's Theorem we have that a.e. => a. unif (so also a.e. => meas), and by Exercise 3.2.5, we also have unif => mean. In this case, we have the chart:

unif

JL(S) <

00

mean.

a. unif

~ meas

a.e.

It is also possible to consider the case where a sequence of measurable function is dominated by a l1-integrable function g, i.e., Ilkl ::; 9 l1-a.e. In this case it follows from the DCT that a.e. mean and, similarly, meas => mean (Exercise 2). We refer the reader to Munroe ([Mu] p. 237) for a complete discussion.

{fd

Exercise 1. Show that a. unif => meas. Exercise 2. Show that the DCT is valid if convergence l1-a.e. is replaced by convergence in l1-measure. [Hint: Theorem 3.6.3.] Exercise 3. Show LO[a, b] is separable.

114

CHAPTER 3. INTEGRATION

Exercise 4. Show a E-measurable function f is p-integrable if and only if :3 a sequence of p-integrable E-simple functions {
MIKUSINSKI'S CHARACTERIZATION - LEBESGUE INTEGRAL

3.B.

3.8

115

Mikusinski's Characterization of the Lebesgue Integral

In this section we give an interesting characterization of the Lebesgue integral due to J. Mikusinski. The characterization is used later in the proof of Fubini's Theorem. Let S be a semi-ring of subsets of S and let p, be a premeasure on S. Let p, also denote the countably additive extension of p, to the .,.-algebra, M, of p,*-measurable subsets of S (§2.4). For Mikusinski's description of the Lebesgue integral, we require two lemmas.

Lemma 1 Let E E M be p,-null and p,(Jk )

<

€

and

f

CA(t)

00

€

> O.

There exists {Jk} C S such that

for tEE [so t belongs to infinitely many Jk for

k=!

tEE}. Proof: For each i, there is a sequence {Iij : j E N} C S covering E such that

f

p,(I,j) < 13/2' (Definition 2.4.1). Now arrange the double sequence {Iij : i,j EN}

j=1

into a sequence {Jd. If tEE, t belongs to infinitely many {Jd so {CJ.(t)} is 1 infinitely often and co

00

00

LP,(Jk)=LLP,(Iij) <e. k=!

We say that a series

;=1 j=1

I: ak is absolutely convergent to

x if the series is absolutely

00

convergent and

I: ak

x.

k=!

Lemma 2 If f : S - t R* is p,-integrable, there is a sequence of S-simple functions {ak} such that the series I: ak is absolutely convergent to f p,-a. e. and k

Proof: By Theorem 3.5.3 there exists a sequence of S-simple functions {'Pd such that II'P.I: fll! - t O. Then 'Pk - t f p,-measure (Proposition 3.7.1) so by Theorem 3.6.3 there is a subsequence, which we continue to denote by {'Pk}, which converges p,-a.e. to f. We may additionally assume that II'PHl - 'Pk III < 1/2k. Set 'Po == 0 and ak

'P.I: -

'Pk-I for k

?: 1. Then

ak

= 'Pn

f Is lakl dfl < 00, the MCT implies that .1:=1 I: ak is absolutely convergent to f fl-a.e. Theorem 3 f : S {1/Jk} satisfying

-t

R is p,-integrable

¢:>

-t

f p,-a.e. or

f

ak

=

f p,-a.e. Since

.\:=1

10'.1:1

converges in R p,-a.e. so the series

there is a sequence of S-simple functions

CHAPTER 3. INTEGR.ATION

116

"IN t)

(2) fit)

for ony t for whIch

t

<

CXJ.

k",]

In theis

COSf,

Proof:

¢=:

The MeT impli('o that.

L

converges p-a.e. ill R. By hYPoUlPsis, f(1)

L

JHU'., the DeT implies t.hat

!

is p.-integrable so the

=

serie~

L:

',bdl) for Sllch Jloints. Since

is fI·-intpgrablc and

IS" fdp =

f

If I

I" l/'kdp.

k==l c}:

Let {ad be

ill Lemma 2 and let E be a It-null set where

f

1(,"(1)1 <

X)

k=l

and set

f (t) = ('.J..

so

then

t>t

!/:

g

Let

:-h} 1)(' a.s ill Lemma I (wit.h respect. t.o

E) and

Define a sequeuc<' oj" S-simple funct.ions, {
J~, 1he series

If I 30,

n..{ l) for t

{h( l) <

L

CXJ

I
S('fi('S

actually only has a.

infinitely oft.pn. II' finit.(,

Humber of 1l00HI('J"O terms]

E and

Lemmas I and 2. ;--';ole, ill particular, I.hat if conditions (1) and (2) are satisfied, the function f io measurable and the series L: converges to I ji-a.f'. One of the important features of" Mikusinski's charac1.erizittion of t.he iutegral is that null seb are not meut.iOlwci. \Ve use t.his to good advantitge in OUf proof of Fubini's Theorem in §3.9. The characterization of the integral in Theorem 3 was given for the Ldwsgue integral ill R \la.cNeille ([Mac]) and for the integral ill R" ,1. Mikminski ([Ml)). Tn Rn the characterization has an even simpler form; the S-sirnple fll11Ctions ill Theorem 3 can be replaced scalar multiples of characteristic functions of half-closed intervals. III this case the value of the integral only depends on the value of the measure of such half-closed intervals so a definition of the Lebesgue integral can be based on the characterizittion which does not require t,he full development of properties of Lebesgue mt'asure. Such a developmt'nt of the integral is carried out in [M2J and [DM].

3.8. MIKUSINSKI'S CHARACTERIZATION

LEBESGUE INTEGRAL

117

Exercise 1. Assume J.I. is complete. Show that if {the} is a sequence of J.I.-integrable functions satisfying (1) and (2), then f is measurable, J.I.-integrable and

118

3.9

CIIAPTER 3. INTEGRATION

Product Measures and Fubini's Theorem

Let (5,S,,..) and (T,T,v) be measure spaces. If A c 5, BeT, a set of the form A x B is called a rectangle in 5 x T and A and B are called its sides. If A E S, BET, A x R is called a measurable recta.ngle. Let R be the family of all measurable rectangles in 5 X T. Since (A x B)n(C x D)

= (AnC) x (RnD)

and

5 x T\A x B

(k x B)U(A x RC)U(N x Be),

R is a semi~algebra of subsets of S x T. We define the prod uet of f1 an d v, denot.ed by Jl x v, on R by seLti ng It ,..(A)v(B) [here O· 00 = 0 as usuall.

X V

(A x 1l)

Theorem 1 ,.. x v is a premeasun; on R. 00

Proof: Let {Ai x B,} c R bc pairwisc disjoint with A x /3 = U A, t::::::l

Then

X

Fl i

R.

00

CAxB(S,l) = LCA,XB,(S,t) for s E 5, t E T. Fix s and integrate with respect to v to obtain from the MCT thai.

CA(s)v(B)

L CA,(s)v(Bi ). 1=1

Now integrate with respect to II and apply the MCT to obtaill 00

It(A)v(B)

= L,..(Ai)v(lJ,) i=]

as desired. The premeasure Ii v can now be extended to a complete measure Oil the o--algebra of (II x v)* -measurable subsets of 5 x T (§2.4). We denote this measure by II x v and call it the product of Ji and v [see, however, Example 2 below]. We refer to the clements of the o--algebra of (Ji x v)" -measurable sets as Ji x v-measura.ble sets and functions which are measurable with respect to this o--algebra as being Ji x v-measurable. Note that if both Jt and v are finite (o--finite), then,.. x v is finite (o--finite). Thus, if Ji and v are o--finite, Ji x v is the unique mea.9ure on the smallest o--algebra containing R, denoted by o-(S x T), which satisfies Ji x v(A x B) = Ji(A)v(B) [Theorem 2.4.121. If m is Lebesgue measure on R, then m x m is a complete measure 011 a o--algebra in R2 which contains all of the measurable rectangles and, hence, contains all of the open (Borel) sets in R2. Thus, m x m and m2 agree on B(R2) and, hence, must be identical (Theorem 2.4.21 and Corollary 2.5.4 (vi)).

3.9. PRODUCT MEASURES AND FUBINl'S THEOREM

119

Example 2 (Bartle) In general the product measure may not be the only countably additive extension of p x v to the a--algebra generated by R.. Let p : peR) --) R* be defined by peA) = 00 if ,4 is uncountable and peA) = 0 otherwise. Let L be the a--algebra generated by peR) x peR). Let (Ps ) be the projection of R x R onto the first (second) coordinate; t) s (PT(s, t) t). Define 7f on L by neE) == 0 if E = G u H, where and PT(H) are conntable, and 00 otherwise. Then 1< is a measure Oil L such that 7f(A x B) = p(A)'l(E) p x IL(A x B). Define p on L by peE) 0 if E = G u II u K, where PT(H) and the projection of K onto the diagonal {(x, x) : x E R} are and p( E) 00 otherwise. Then p is a measure on L such that peA x B) p(A)f1(B). If E y) : x + y = O}, t.hen EEL but pee) 0 while 7f(E) 00. Many other pathologi<'s can occur for product m<'asureSj s<'e, for example, :Ga] or

[DC] Vve now consider the evaluatio'n of integrals with respect to a product measure. As is the case with the Riemann integral, the most efficient way to evaluate an integral with respect to a product measure is as an iterated integral. A result which a.sserts the equality of an integral with respect to a product measure and an iterated integral is often referred to as a Fubini theorem. For example, if f : 51 x T .. ) R is R.-simple and 11 x il it follows from the definition of the product measure that

r

isxT

fdp x

il

r[

rr

f(s, t)dll(B)dv(t) = f(s, 1r./s·" is iT

=

I.e., Fubini's Theorem bolds for such functions. \rYe now use Mikusinski's characterization of the integral to ext<'nd Fubini's Theorem to Ii x iI-integrable fUIlctions. Ifenc(>forth, we assume that the measures p and v aTe both complete. If f ' S x T ..... R*, for .'i E S (t E T) we denote by ,) (f(', t)) the function f(8,·) : 1' ..... R* (fe t) : S ..... R*) defined by feB, .)(t) t) (f(·,t)(s) = f(s, t)), \Ve follow the usual procedure in Fubini's Theorem of agreeing that if a function 'P is defined IJ.-a,e, in S (v-a.e, in T), lhen 'P (1,':') is extended to all of SeT) by setting it equal to 0 OIl lhe p-null (vnull) set where it is undefined. Thus, if {'Pk} is a sequence of S-measurable functions which converge pointwise p-a.e. in S, we may assume that there is an S-measurable functioll 'P defined 011 S such that. 'Pk ..... 'P 11-a.(>. [t.he completeness of IL is used here]. This situation is encount.ered several times in the proof of Fubilli's Theorem.

Theorem 3 Let

f :S x T

..... R be p x v-integrable. Then

(i) f(s, ,) is v-integrable for p almost all s E S, (ii)

.5 .....

(iii) IsxT fdp x

, )dv is 11 il

1:9

I.THPf11·f1.fH

and

·)dvdp(B) = fsfTf(B,t)dv(t)dp(s).

CHAPTER 3. INTEGRATION

120

Proof: By Theorem 3.8.3, there is a sequence {v';d of S x T-simple functions

IsxTIv';kldlL x v <

such that

f

the series

k=1

00

f

and f(s,t)

k=1

v';k(S,t) for all (s,t) for which

lv';k(S,tll converges and

r

isxT

fdJL x v

=

f r

k=lisXT

v';kdJL x v.

(3.1)

By the MCT

E

isXT IIPkldJL x v

=

(3.2)

k:1 is£ lv';k(S,t)ldv(t)dJL(s)

r f r lv';k(S,t)ldv(t)dJL(s) <

is k=1 iT

00

'I

E,

!f'k(S, t) is absolutely convergent in R for v-almost all t E T. [The v-null

sel,

so there is a JL-null set E such that

IT lv';k(S,t)ldv(t) <

00

for s

'I

E. For s

the MCT implies that

so

f

k=1

may depend on s.] For such s, t, f(.~, t)

f(·~,·J

f

k=!

=

f

k=!

v';k(S, t); in particular, for

S

'I

E,

v';k(S,') v-a.e., and since

the DCT implies that f(s,') is v-integrable and IT f(s, ·)dv = ther, for s

'I

E,

it r v';k(S, ')dvl:5: lk=lh

co

r lv';k(S, ')1 dv

h

and the function on the right hand side of this inequality is Ii-integrable by (2) so the DCT implies that S -> IT f( S, . )dv is p-integrable with

ish f(s, ·)dvdIL(.»

f r rv';k(s,t)dv(t)dp(s) = isrxT fdp

k=1 is iT

X

v

by (1). The main difficulty in applying Fubini's Theorem is verifying the p x v-integrability of the function f. A result which is often used for this is a result called Tonelli's Theorem.

3.9. PRODUCT MEASURES AND FUBINI'S THEOREM Theorem 4 Let J1., v be a-finite and f : S x T measurable. Then

(i) f(s,') iB T -rneaB1tmble for

(ii) s

->

fr

Il-ulmo.~t

->

121

R* non-negative and Jl x v-

all s E S

f(s, t)dv(t) i$ S-1fL/:U:5'Urable

(iii) ISxT f dJl x v

Is IT f(8, t)dv( t)dJ1.(s) [the integrals might be

[Of course, there is an analogous statement with the variables

5

and t reversed.]

Proof: Since Jl and v are a-finite, Jl x v is a-finite so there exists an increasing sequence of Jl x v-measurable sets {E;J such that Ei j S x T and Jl x v( Ei) < 00. Set f. = (J 1\ i)CE •. Then each fi is Jl x v-integrable. By Theorem 3, 1;( 5, .) is v-integrable for Jl-almost all s E S and since f,(8,.) j f(8,·) on T, f(8,') is T-measurable for Jlalmost all 8 E S and by the MeT,

fr

fi(5, t)dv(t) j

fr

f(8, t)dv(t) for Jl-almost all

5

E S.

(3.3)

By Theorem 3, the function .~ fr fi(8, t)dv( t) is S-measurable so by (3) -> IT f(8, t)dv( t) is S-measurable and the MeT applied to (3) implies that

8

r r 1;(8, t)dv(l)dJl(s) T ~h· r r f(8, t)dv(t)dJl(8).

~~ Since fi

(3.4)

T f on S x T, the MeT implies

r

isxT

fidJl x v T

r

isxT

f dJl x

V.

(3.5)

By Theorem 3,

80

(4) and (5) imply

Tonelli's Theorem can be used to check the integrability of a measurable function with respect to a product measure. If Jl and v are a-finite and f : S x T R is Jl x v-measurable, we can apply Tonelli's Theorem to the function Ifl to check its Jl x v-integrability. If If I is Jl x v-integrable, then Fubini's Theorem can oftcn be applied to evaluate the product integral IsxT f dJl x v. As an application of Tonelli's Theorem we show how the product measure of a set can be computed as an integral. If E C S x T and s E S, the s-section of E at 8 is defined to be E. = {t E T : (8, t) E E}. Similarly, if t E T, the t-section of E at t is defined to be Et = {s E S : (5,t) E E}. We have the following elementary proposition.

CHAPTER 3. INTEGRATION

122

Proposition 5 Let E C S x T and

S

E

S. Then

(i) GE(S, t) = GE.(t), (ii) (Ee).

(E.)C,

(iii)

(Y E.).

U(Eo)., where

(iv)

mEo). =

n(E.)., where Eo C S

o

a

c

S x T, X

T.

•

From Proposition 5 and Fubini's theorem, we have Theorem 6 Let E C S x T be Jl x v-measurable and have finite It x v mCa.'mre. Then

(i) for Il-almo"t all (ii) the junction s (iii)

$

E

S the .lcction Es E T,

-+ I/(E,)

Is v(Es)dll(S) =

is /l-intcgmble, and

It x veE).

Remark 7 If It and v arc O'-finife, by Proposition 5 and Tonelli'8 Theorem conclitiolls (i), (ii)' s -+ v (E,) is S-measurabl~ and (iii) also hold for any E C S x T which is Jl x v-measurable. Remark 8 In some of the deVelopments of Fubini's Theorem for product 11Iea"ures, Theorem 6 is an important intermediate step in that it establishes the result for integrable functions. The general result is then established by approximat.ing the general integrable function by simple funcLions. The proof of Theorem 6 from basic properties of product measures can be somewhat lengthy (see [Roy] 12A.Lj·18). We now give several examples which illustrate the potheses in the Fubini and Tonelli Theorems.

for the various hy·

Example 9 Let S = T = [0,1]' Jl = Lebesgue measure and 11 counting measure restricted to the Lebesgue mea.sura.ble subsets of S. Set E = {(x, x) : 0 :<:: x :<:: l}. Note that E is It x v-measurable. Then

while

II II

GE(x,y)dm(x)dv(y)

0

GE(x,y)dv(y)dm(x) = 1.

This example illustrates the importance of the O'-finiteness condition in Tonelli's orem [also in Theorem 6 and Remark 7].

The~

3.9.

PRODUCT MEASURES AND FUBINl'S THEOREM

°

y)

Example 10 Let /(0,0) . C" iJ Y <1 _ , 0 tlterWlse. ,,!nce &;;

°

1111 o

Sirnilarl.Y,

(;y2 _

= J(x,y) for x

.

/(x,y)dyd:L" =

0

123

11 0

tt

1 -.--:--d:I: 1 + :r2

io io f (x, y)duiy

for

i- 0,

°

= 7rj-1.

-n/4.

This example shows that the non-negativity condition important. A similar examp]<' given in Exercise 2.

III

Tonelli's Theorem

IS

Example 11 (Sierpinski) [Sir]. This example points out the importance of the med.surabiliLy aSSllmptiOil in T
Jd Id

Jd Id

I\ote thai eiJch f(.r,·) (f(.,y)) is it I~o[{'l function but..r is not Illeasurabk, i.e., J is "separately llleasurable" but. not "jointly Ilwasurable". Another iuterestillg exaIllple rdative to it.erated integrals is in Exercise :J. There is a.nother stalldard nwlhod of constructing product Illeasures whirh we !lOW ollL]im:. 'vVe assume that // and // arc cr-finite. If c' cr(S x T), then for .s E S (/ E 'r) every section Es T (El E S) il.lld t.he functioll S -'> I/(I~;s) (t -'> /1([;;/)) is S-llIca:;urable (T-lIlcilsurab]e). The set flilictioll A defilled on cr(S x T) by

is it iT-finite measure which satisfies A(A x B) 1'(1\)1/(8) when A S, BET. (uHI is defiw·d to be the product of I' ilnd //. In genC'ral, this measure is not complete. TIl particular, if /1. = /1 Lebesgue n1PaSllre Oil R, then A is not complete and 2 'lO is not two-dimensional Lebesgue Hlcasur(' Oil R which is sOIIlcwhat annoying. Fubini's ThcO[(,m is llwlI (,stablished by approximating functi()!Js simple functions. Sec, for example. the devel0Plllent in [Hal]. Fubini's Theorem fails miserably for finitely additive set see [HY1.

Exercise 1. Let 9 : S R (h : l' -'> R) be S-measurable (T-lIleil.sllrablc) and dcflIle f: S x T .-> R by .f(s, t) g(s)h(t). Show J is iT(S x T)-measurable. If 9 and h a.re integrable, show

f

is integrable and

[ frip x ~xT

/1

= [ 9
iT

124

CHAPTER 3. INTEGRATION

Exercise 2. Show I; It';:'(e~:r;Y-2e~2xY)dxdy and It" I~(e~:r;Y-2e~2xY)dydx both exist but are not equal. Exercise 3. Let f(O,O)

0 and f(x,y)

llll but

f

= xy/(x 2 + y2)2 otherwise.

f(x,y)dxdy

l1l1

Show

f(x,y)dydx

is not (Lebesgue) integrable over [-1, I] x [-1,1]. Hint: Consider [0, IJ x [0,1].

Exercise 4. Give an example of a set in the plane which is not Lebesgue measurable. Generalize to Rn. Exercise 5. Show that the conclusion (i) of Theorem 6 cannot be improved to: every section E. E T. Exercise 6. If E E u(S x T), show every section E. E T. [flint.: Consider all E C S x T such that E. E T] Exercise 7. Let It, v be u~finite. If E, F are It x for It~almost all 8 E S, show It x v(E) It x v(F).

v~measurable

and It(E.)

= v(F.)

Exercise 8. Show u(8(R) x 8(R» = 8(R2). [Hint.: Consider the projections from R2 into the coordinates and use Exer. 2.5.11.]

Exercise 10. If f : S x T --+ R is u(S x T)~measurable, show f(8,·) is for every 8 E S. Docs the same hold for It x v~measurable functions?

T~measurable

x1 Exercise 11. Evaluate Iooo e- dx = / by writing /2 as a double integral in the plane and using polar coordinates.

Exercise 12. Let

f : Rn

--+

R be Lebesgue integrable. Show

r

JRn

f(x+y)dx

r f(x)dx JRn

for every y ERn. What is the corresponding formula for IRn f(ax)dx, a E R?

3.9. PRODUCT MEASURES AND FUBINI'S THEOREM Exercise 13. For p > 0, q > 0, the Beta Function B(p, q) is defined by

B(p, q) Show B(p,q)

=

l

(1 - x )q-1dx.

B(q,p) and r(p)r(q) = f(p+q)B(p,q).

Exercise 14. Evaluate f~

J;

y2 e

dydx.

125

126

3.10

CFIAPTER 3. INTEGRATION

A Geometric Interpretation of the Integral

We show that Fubini's Theorem can be used to show that the integral can be interpreted to be the "area under the curve". Let, (S, S, 11) be a a-finite, complete measure space. If f: S -+ R, the ordinate set of f is rl j = ((x,y): x S, -CX) < y < f(x)}. Proposition 1

f

is S -measurable ~ rl j is 11 x m-measurable.

Proof; {=: For y E R, rl~ = {x E S: (x,y) E rl j } = {x : f(x) > y}. rlj E S for m-almost all y E R (Remark 3.9.7). Thus, rlj E S for y in some dense subset DC R. Therefore, {x: f(x) > r} E S for rED and f is S-measurable by 3.1.L =}: Suppose f : S -+ R is S-measurable. For r E Q let

A, B,

= {y

{xES:f(x»r}, E R:

-CX)

< y ::; r},

and E

= U{ A,

X

B, : r E Q}.

Then E is It x m-measurable, and we claim that E rl j . First, suppose that (x,y) E E. Then (x,y) E A, X B, for some r E Q so -CX) < y ::; r < f(x) and y) E rl j • Hence, E C rl j . :\'ext, suppose y) E rl j . Then -CX) < y < f(x) so there exists r E Q such that -CX) < y ::; r < f(x) and (x,y) E A, x Br C E. Hence, rl j CEo Theorem 2 Let

f :S

R be non-negative and S -measurable, and let H

{(x,y):xES,O::;y
Then H is 11 x m-measurable and 11 x m(H)

Is fdlt.

Proof: H is measurable by Proposition 1 since H 3.9.7,

11 x m(H)

= Hj\S x

0). By Remark

= ism(IF)dll(x) = isf(x)dll(x).

If the basic properties of the product measure are developed before the int,cgral is defined, the com,:lusion in Theorem 2 can be used to define the integral for nonnegative measurable functions. This approach to the integral leads to some very quick proofs of the convergence theorems, particularly the MGT. For a description of this method of defining the Lebesgue integral see [Zl].

Exercise 1. If f is S-measurable, show the graph of f, G 11 x m measure O. Exercise 2. Show is fdll

=

1"0 11 ({ x: f(x) > y}) dy.

{(x, f(x)) : xES}, has

127

3.11. CONVOLUTION PRODUCT

Convolution Product

3.11

As another application of Fubini's Theorem we consider the convolution product of two functions. The convolution product has many important applications in analysis, from approximation theory to integral transform theory. We give two such applications in Theorem 11 and Exercise 2. If f, 9 : Rn ---> R, the convolution product of f and g, denoted by f * g, is defined by

f*g(x)=

f JRn

f(x-y)g(y)dy,

provided the integral exist.s. We use Mikusinski's characterization of the Lebesgue integral to show that the convolution product of two Lebesgue integrable functions exists a.e. and defines a Lebesgue integrable function. Other conditions which guarantee the existence of the convolution product are given in the exercises. Theorem 1 Let f, 9 E LI(Rn). Then

(i) f (ii) f (iii)

* g(x) *9

E

=

JRn

f(x - y)g(y)dy exists for m-almost all x ERn,

LI(Rn),

Ilf * gill :s Ilflll IlglII·

Proof: We use Fubini's Theorem to show that the function

H(x,y) = f(x - y)g(y) is Lebesgue integrable over R2n and then (i) and (ii) will follow from Fubini's Theorem. Let {f;} ({g;}) be Lebesgue integrable Borel functions such that

Lllf.11 1 < 00

(Lllg.III < 00) i~l

i=l

and

f(x) = Lf.(x)

(g(x) = Lgi(X»

i=l

i=l

for any x for which the series is absolutely convergent (Theorem 3.8.3). Set F.(x,y) = f.(x - y), G.(x,y) = g.(y) and note each Fi , G i is a Borel function on R2n. 00

Consider the Cauchy product,

j

L L. FiG j - i , of the two series L Fi and L G •.

j=l i=O

Since

CHAPTER 3. INTEGRATION

128

I: ;

~ I~ fIt2n FiG j_, ~ ~ fItn Ifil fItn Ig)-i I < DO, this last series being the Cauchy product of the series L:

Ilfdl]l L: Ilgi Ill. ][ [or

some

(x,y) E R2 the series j

00

LL

Fi(X, y)Gj_i(x, y)

j=1 ;=0

converges absolutely, then both series

I: f~(x,y), I: G,(x,y)

must converge abso-

1=1

t=]

lutely ([Kn] Theorem 11, p. 89) and, therefore, must converge to F(x, y) and G(x, y), respectively, so 00

)

LL

Fi(x, y)G)_i(X, y)

)=1 i=O

must converge to F(x,y)G(x,y) = H(x,y) ([Kn]). By Exercise 3.8.1,11 is Lebesgue integrable over R2n, and (i), (ii) follow from Fubini's Thoerem. For (iii),

fRn

If

* gl < fItn fRn If( T - y )g(y) Idydx fItn fItn If(x - y)g(y)1 dxdy fItn Ig(y)1 dy fItn If(x)1 dx

by Tonelli's Theorem. Note that the proof of Theorem 1 shows that the [unction (x,y) ---> f(x - y) is Lebesgue measurable (see Exer. 8; [or another proof o[ the measurability see [I1S] 21.31 ). Several of the algebraic properties of the convolution product are given in the exercises. We now show that the convolution product can be used ill establishing approximation results.

Definition 2 A sequence {'Pd 8-sequence if

c

LI(Rn) is called an approximate identity or

(i) 'P k ( x) ?:: 0 fa r all x ERn, (ii)

II'Pklll

= 1 for all k,

(iii) for every neighborhood of 0, U, in Rn, limfu< 'Pk(x)dx = O. k

We give examples of 8-sequences.

Example 3 In R, take 'Pk(t) = ~C[-l/k,l/kl. Example 4 'Pk(t) =

~I+tkt)2' t

E R.

Example 5 In R, 'Pk(t) = Cke-kt2 where Ck is chosen such that (ii) holds.

(J

3.11. CONVOLUTION PRODUCT

129

We denote by C~(Rn) the vector space of all functions '-P on Rn with compact support which have continuous partial derivatives of all orders. It is not apparent that there are any non-zero functions in C~(Rn), but examples of such functions are given in the examples below. 2

Example 6 In R, pick '-P E C~(R), '-P :::: 0, [like '-P(t) = e- I !(I-t ) for It I < 1 and '-P(t) = 0 for It I :::: 1]. Set '-Pk(t) = ck'-P(kt), where Ck is chosen such that (ii) holds. Note that if'-P is chosen as above spt('-Pk) C [-11k, 11k].

Example 8 In Rn, pick '-Pk as in Example 6 and define ,pk as in Example 7. Note in this case, 1/;k E C~(R") and spt(1/;k) C {x: IXkl ~ 11k}. Example 9 Define'-P on R" by '-P(x) = e-I/(l-lIxW) if Ilxll < 1 and '-P(x) = 0 otherck'-P(h) where Ck is chosen to satisfy (ii). Then '-Pk E C~(R") and wise. Set '-Pk(X) spt('-Pk) C {x: Ilxll ~ 11k}. We next establish an intermediary result which is interesting in its own right. Theorem 10 Let f : Rn --t R* be Lebesgue integrable and for h E Rn define fh : Rn --t R* by fh(x) I(x + h). Then lim Ilfh - fill O. h-+O

Proof: Let E > O. By Theorem 3.5.6 there exists '-P E Co(R") such that III - '-Pill < f. There exists a > 0 such that '-P(x) 0 for IIxll > a. '-P is uniformly continuous so there exists 1 > 8 > 0 such that I'-P(x) '-P(Y) I < c when IIx - yl/ < 8. Thus, IIhl! < 8 implies

f 1'-P(x+h)-'-P(x)ldx
JRn where A

= m{x : IIxll II/h -

1111

~

a + I}. Thus, if Ilhl! < 8,

~ Ilfh

'-Phlll

+ II'-Ph -

'-PIlI

+ II'-P

fill <

t

+t

.

A

+ t.

Theorem 11 Let {'-Pd be an approximate identity in LI (Rn). Then for every IELl(Rn),lirnll/*'-Pk III1 O. Proof: Let c > O. By Theorem 10 there exists 8 > 0 such that IIhll ~ 8 implies !Ifh - 1111 < €, where Ii.(x) I(x + h). By (iii) there exists N such that k :::: N implies ~lxll::>:6 '-Pk( x )dx < (.. Then IIf * '-Pk - fill

fRn IJRn (f(x - y) - f(x))'-Pk(y)dyldx fRn '-Pk(y)JRn I/(x - y) - l(x)1 dxdy JRn '-Pk(Y) II/-y - fill dy Jllyll9 '-Pk(Y) II/-y - fill dy + ~IYII>6 '-Pk(Y) Ilf-y - fill dy < d llyll9 '-Pk(y)dy + 211/111 JIlYII~6 '-Pk(y)dy < € + 211fl!1 f. ~

CHAPTER 3. INTEGRATJON

130 Corollary 12

C~(Rn)

is a dense subset

0/ VeRn)

(with respect to 11111)'

Proof: By Theorem 3.5.6 it suffices to show that C~(Rn) is dense in Cc(Rn). Let WE Cc(Rn) and K = spt(W). Choose {'Pk} C C~(Rn) as in Example 9. If x E Rn and dist(x,K) > 11k, then 'Pk(X Y) 0 for y E K so

vanishes for such x, i.e., spt('Pk * W) c {x : dist(x,K) ~ 11k}. Hence, 'Pk * Whas compact support and by Exercise 3 'Pk * Wis infinitely differentiable so 'Pk * WE C~(Rn). By Theorem 11, II'Pk * W will - t O.

Proposition 13 Let KeRn be compact and V :) K open. Then there exists WE C~(Rn) such that 0 ~ W~ 1, W 1 on K and W= 0 on ve. Proof: We may assume that V is bounded. Let 0 = dist(K, ve) > 0 and let Kl be a "o/3-neigborhood of K", i.e., Kl {x: dist(x, K) < o/3} [a sketch at this point is useful]. If k is chosen such that 11k < 6/3, the function W= CK] * 'Pk is in C~(Rn) if 'Pk is chosen as in Example 9. Clearly 0 ~ W~ 1. If x E K, then (Exer. 1)

since the integration is over a ball of radius < 6/3 and center x. If x W(x) = 0 since the integrand in the convolution product is O.

f/.

V, then

Exercise 1. If /, g, h E LI(Rn), show / * 9 = 9 * f, f * (g * h) (f * .lJ) * h, and f * (g + h) = f * 9 + f * h. Thus, VeRn) is a commutative algebra with convolution as product. Show that it does not have an identity.

Exercise 2. If f E LI(R) and f(t) 0 for t < 0, its Laplace tranBform is defined by C{f}(s) fooo e-'!f(t)dt. Show this integral exists for s :::: O. If f, 9 E VCR) vanish for t < 0, show f * 9 vanishes for t < 0 and C{f * g} = C{f}C{g}. Exercise 3. If f E C~(Rn) and 9 E VeRn), show f * 9 E Ll(Rn) give a formula for calculating the partial derivatives of / * g.

n COO(R) and

Exercise 4. Let {'Pk} be as in Example 9. If f : Rn - t R is bounded, continuous, * 'Pk is uniformly continuous and f * 'Pk - t / uniformly on compact subsets of

show f Rn.

Exercise 5. Let f E Ll(Rn). If fRn f'Pdm a.e.

0 for every 'P E C~(Rn), show f

0

3.11. CONVOLUTION PRODUCT

131

Exercise 6. Let IE V(Rn) and g : Rn -+ R be bounded, measurable. Show exists everywhere in Rn and is uniformly continuous. Exercise 7. Let {'Pd be an approximate identity. If g measurable and continuous at x, show 'Pk * g(x) -+ g(x). Exercise 8. Show that if I(x - y) is measurable.

I : Rn

->

-+

R is bounded,

R is measurable, then the function (x,y)

Exercise 9. Let E C R be Lebesgue measurable with m (E) contains an interval. Hint: Assume E C (a, b] and define

I(x)

Rn

1* g

> O. Show that E - E

tCdy)CE(X+y)dy.

Show I is continuous at 0 and 1(0)

> O. Consider {x: I(x) > OJ.

Exercise 10. If I E C C (Rn) and {'Pd is an approximate identity, show {J converges to I uniformly on R.

* 'Pd

+ spt (g»

and

Exercise 11. If I, g E C c (Rn), show spt (f

I*gECc(Rfi).

* g)

C dosure(spt (f)

CHAPTER 3. INTEGRATlON

132

3.12

The Radon-Nikodym Theorem

\Ve now consider one of the most important results in the theory of measure and integration, the Radon-Nikodym Theorem. The Radon-Nikodym Theorem essentially as an indefinite integral. characterizes the integral as a set

Definition 1 Let p, v be signed measures on the u-algebra Then v is absolutely continuous with respect to p, denoted by v if IJlI (E) = 0 implies Ivl (E) = 0 whenever E E 2::. Example 2 If p is a measure and f : S R+ has a Jl-integral and v(E) = IF: fdJl for E E 2::, then v
Proposition 3 Let p, v be signed measures on

(i) v
v is said to be p-continuous if

I!JI(E)~O

By Theorem 3.2.17 indefinite integrals of Jl-integrable functions are Jl-continuous. Concerning absolute continuity and p-continuity, we have Proposition 5 (i) If v is p-continuous, then v
¢=

follows from (i).

¢:}

v is Jl-continuous.

Suppose not. Then there exists (. Set E lIinE) =

( > 0 and {Ej} C 2:: with Ipl (Ej ) < 1/2 j and Ivl (Ej ) ~

= n U Ej. k=lj=k 00

Then for each k,

Ipi (E) ~

I: Ipi (E

j )

j=k

~

l:n

3.12. THE RADON-NIKODYM THEOREM so IfLl (E)

= O.

But, Ivl finite implies Ivl (E)

= lim Iv I (U k

i=k

E j ) ? rim Ivl (Ed:;;' { so

V

is not absolutely continuous with respect to fL. The finiteness condition in (ii) cannot be dropped even when fL is finite.

L

Example 6 Define fL, von P(N) by fL(E)

2k. Then v
kEE

but v is not fL-continuous. For the proof of the Radon-Nikodym Theorem we first require a lemma. Lemma 7 Let fL, v be finite measures on L with v 0 and A is a (v - cfL)-positive set.

{ > 0 and A

Proof: For each k let (Pk , N k ) be a Hahn Decomposition for the signed measure v - tfL. Set Ao =

UP

k

k=1

and Bo =

n

k=1

N k • Since Bo C Nk for every k,

so v(Bo) O. Thus, v(Ao) > 0 and since v ~ fL, fL(Ao) > O. Hence, fL(Pk) > 0 for some k. For such a k, set C 11k and A Pk . We are now ready for the proof of the Radon-Nikodym Theorem. We say that a signed measure v is O'-finite if Iv I is O'-finite. Theorem B (Radon-Nikodym) Let fL(v) be a O'-finite (signed) measure on L with v
and set

A = SUP{h9dIL : 9 E F} ::; v(S) <

00.

Choose a sequence {fd C F such that limIs fkdfL A. We construct an increasing sequence from {fd by setting gk II V ... V /k. Then gk i and gk E L1(fL). Moreover, gk E F since for any EEL if EI {t E E : !t(t) = gk(i)} and

Ei+J

{t E E: fHI(t)

= gk(t)}\Ei for

1 ::;

i::;

k - 1,

CHAPTER 3. INTEGRATION

134 then

Also). 2: Is gkdJ.l 2: Is fkdJ.l for every k implies lim Is gkdJ.l =).. By the MeT if f = limgk, then IsfdJ.l =). < 00 so f is finite J.l-a.e. and we may assume f E £1 (J.l). Since gk E F, f E F by the MeT. We claim that v(E) = IE fdJ.l for E E 2:. Let va be the measure defined by vo(E) = v(E) - IE fdJ.l. If va is not zero, then since va « J.l, Lemma 7 implies that there exists f > 0 and A E 2: such that J.l(A) > 0 and A is (va - fJ.ll-positive, i.e.,

f fCAdJ.l

lE

Thus, if h =

= fJ.l(EnA):-:; vo(EnA) = v(EnA) -

f + fC A ,

f hdJ.l

lE for every E E

=

f

lEnA

fdJ.l for E E

E.

then

f fdJ.l + fJ.l(E n A):-:; f

lE

lE\A

fdJ.l

+ v(E n A)

:-:; v(E)

2: so h E F. But,

is

hdJ.l =

is

fdJ.l

+ fJ.l(A)

which contradicts the definition of ).. Hence, Uniqueness of f follows from 3.2.18. Assume that J.l and

V

V

= ).

+ fJ.l(A) > ).

= fdJ.l.

are O"-finite. Then S =

UE j , where Ej E 2:,

j=1

E j C EJ +1

and J.l(E j ) < 00, v(Ej) < 00. By the first part, for each j there exists a non-negative, J.l-integrable function gj such that gj = 0 on EJ and v(E n Ej l = Ie gjdJ.l for E E 2:. From the uniqueness of gj, it follows that gj = gj+k J.l-a.e. in E J for k 2: 1. Set Ii = max{gl' ... ,gj} so {Ii} is increasing. Then

Set

f

= limfJ. From Proposition 2.2.5 and the MeT,

v(E)

= limv(E n E j ) = lim hIidJ.l =

h

fdJ.l for E E

E·

Uniqueness again follows from 3.2.18. Note that the function f is not generally J.l-integrable. In fact, f is J.l-integrable if and only if V is finite. The O"-finiteness condition on v can be eliminated [see [Roy] [or a proof]; however, the a-finiteness condition on J.l cannot be dropped as the following example shows.

Example 9 Let S = [0,1] and 2: the O"-algebra of Lebesgue measurable subsets of S. If J.l is counting measure on 2:, then m « J.l but there exists no function fELl (J.l) such that m = fdJ.l [Proposition 3.2.14]. See also Exer. 11.

3.12. THE RADON-NIKODYM THEOREM

135

The function f in the conclusion of Theorem 8 is called the Radon-Nikodym derivative of v with respect to J-I and is denoted by ~~. Note the Radon-Nikodym derivative is only determined up to J-I-a.e. Some of the basic properties of this "derivative" are given in the exercises. The Radon-Nikodym Theorem can be extended to a more general class of measures than the a-finite measures. One important class of such measures are called decomposable measures; for a description of these measures see [HS] p. 317. Necessary and sufficient conditions are known for a measure J-I to admit the conclusion of the Radon-Nikodym Theorem for any signed measure v; such measures are called localizable and have been characterized by Segal [see [TTj for a descriptionj. For other methods of proving the Radon-Nikodym Theorem see [Royj and [R2] and §3.12.1. The Radon-Nikodym Theorem fails for finitely additive set functions; see section 3.12.1 and Exer. 3.12.1.2. Exercise 1. Let J-I, VI, V2 be signed measures on L with one Vi being finite and J-I a measure. If Vi q:: J-I, show (avi + bV2) q:: Ii. If the measures are a-finite, compute d(uv,+bv,) dl'

Exercise 2. Let 5(7) be a a-algebra. Let a, J-I (p, v) be a-finite measures on 5(7). If J-I q:: a, V q:: p, show J-I X v q:: a X p. Compute the Radon-Nikodym derivative of Ii X V with respect to a X p. Exercise 3. Let Ii, v be a-finite measures on f~ E L1(J-I) and J fdv = J f~dli'

L

with v q:: Ii. If

f E

LI(V), show

Exercise 4. Let A, J-I, v be a-finite measures on L with v q:: Ii q:: A. appropriate conditions, show ~ = ~;jX [Chain Rule]. Exercise 5. Let Ii, v be finite measures on 1/ d" dv • J-I-a.e. an d !Y:. dv

L with

Exercise 6. Let v, Vk, J-I be a-finite measures on

L

Under

Ii q:: v, v q:: J-I. Show ~

with v =

f

=I 0

vk. If vk q:: J-I for

k=1

every k, show v q:: Ii and ~

f~

k=1 d" .

Exercise 7. Let A be an algebra which generates If v is J-I-continuous on A, show v < < J-I on

L.

Let Ii, v be finite measures on

CHAPTER 3. INTEGRATION

136

Exercise 8. Let Vk, J1 be signed measures on Then {vd is uniformly J1-continuous if lim Vk (E) 0 uniformly in k. If fk E Ll (J1) and Vk fkdJ1, show {Vk} is uniformly II'I(EJ-O

J1-continuous if there exists 9 E Ll (J1) such that Ifk I S 9 J1-a.e. Exercise 9. Let J1 be a finite measure on Land J1integrable if and only if sup

is

Ifkl dJ1

!k

E L 1 (J1). Show

{fd

is uniformly

< 00

and {fkdJ1} is uniformly J1-continuous. Exercise 10. Let v : L'~ R' be finitely additive and J1 be a mea.sure on J1-continuous, show v is countably addit.ive.

L.

If v is

Exercise 11. Let S R and let L be the
--; f

J1-measure, show

3.12. THE HADON-NIKODYM THEOR.EM

3.12.1

137

The Radon-NikodYIll TheoreIll for Finitely Additive Set Functions

Bochner has given an extension of the Radon-Nikodym Theorem to finitely additive set functions defined on an algebra ([Boll. Bochner's original proof depended on the countably additive version and Stone space arguments. Dubins has given an elementary proof which we present ([D]). Dubins' proof depends on order properties of the space of finitely additive set functions and is independent of the countably additive version of the Radon-Nikodym Theorem. Let A be an algebra of subsets of S. We let ba(A) be the space of all real-valued, bounded, finitely additive set functions defined on the algebra A. ba(A) is a vector space if we set (p+v)(A) = p(A)+v(A) and (tp)(A) ip(A) for p, v E ba(A), t E R, A E A. We write p 2:' v if and only if p(A) 2:' v(A) for all A E A; if p 2:' 0, we say p Ipi (S) for p E ba(A). is non-negative. For convenience of notation, we set Ilpll If p, v E ba(A), the meanings of v 4.( p and v being p-continuous are as in Definitions 3.12.1 and 3.12.4. If f : S -+ R is an A-simple function and 11 E ba(A) is non-negative, then the integral of f with respect to 11 is meaningful and

1£ fdPI ::; sup{lf(t)1 : t E S}p(A) for all A E A [Remark 3.2.2] and A -+ fA fdp defines an element of ba (Alj we denote t.his set function by Bochner's Radon-Nikodym Theorem asserts that if v is Il-continuous, then v can be approximated by an indefinite integral fdlt where f is a simple function. We now prove Bochner's Theorem. We require two lemmas. The first is an interesting result related to the order defined on ba(A) and asserts that any two elements of ba(A) have an infimum with respect to the order defined on ba (A). Lemma 1 Let 11, v E ba(A). Then P 1\ v(A) = inf{p(E)

+ v(A\E)

: E E A, E ~ A}

defines an element p 1\ v E ba(A) which is such that P, v 2:' p 1\ v and if a: E ba(A) and a: ::; 11, a: ::; v, then a: ::; p 1\ v. Proof: Let A, B E A, A n B

p(E)+v((AUB)\E)

= 0.

If E E A, E C AU B, then

= p(EnA)+p(EnB)+v(A\E)+v(B\E) 2:'

so P 1\ v(A U B)

,

2:' p

1\ v(A)

+ p 1\ v(B).

pl\v(A)+pl\v(B)

CHAPTER 3. INTEGRATION

138

Let

6

> 0. There exists El C A, E2 C B, Ei E A such that J.ll\ v(A) > J.l(Ed

+ v(A\El) -

6/2

and so

J.ll\ v(A)

+ J.l/\ v(B) >

J.l(E 1 U E 2 ) + v((A\Ed U (B\E2» J.l(E1 U E 2 ) + v((A U B)\(EI U E 2» J.ll\v(AUB)-6.

6

6

Hence,

J.ll\v(AUB)

J.ll\ v(A)

+ J.ll\ v(B)

and J.ll\ v E ba(A). Taking E A shows J.ll\ v ::::: J.l and taking E = 0 show J.l ::::: v. If a S It and a S v, then clearly a S J.ll\ v. For the next lemma we adopt some unorthodox but useful notation. If 11" v E ba(A) and 6 ::::: 0, we write J.l S v + 6 if f.l(A) S v(A) + 6 for every A E A; note here that J.l, v are set functions and 6 is a non-negative real. Lemma 2 Let

6:::::

0, k> 0, J.l,

E ba(A) with J.l non-negative. If

11)

--kf.l then for every

f.'

f.

<

11)

< kf.l + f.,

> 6 there exists a two-valued A-simple function f such that k , -f.l- 6 < w 2

(3.1 )

and an A-simple function f such that - e;' < w - f df.l < Proof: Choose A E A such that w( A) > equal k/2 on A and -k/2 on N. Then

r

nA) and since weE n AC)

< 1':'

weE

nA

lEnA

11)(

fdf.l S

(:3.2)

6',

E) - (6' - 6) for all E E A. LeL

~J.l(En A) +6, 2

f

(3.3)

1':, C )

r

lEnA'

fdp. < 1':'

-

I':

k

+ -2 J.l(E n A").

The inequality on the right hand side of (1) now follows from (3) and (4). The ot.her inequality in (1) is similar. From (1) it follows by induction that for every j and TT > I': there exists a simple function f such that k

3.12. THE RADON-NIKODYM THEOREM

139

and this gives (2). We are now ready to state and prove Bochner's Radon-Nikodym Theorem which asserts that if II E ba(A) is {t-continuous for some {t E ba(A), then II can be approximated by an indefinite integral with respect to {t of an A-simple function [condition (iii) in Theorem 3 below].

Theorem 3 (Bochner) Let {t, v E ba(A) with It non-negative. The following are equivalent:

(i) II is {t-continuous, (ii) for every IIv -:-

£

> 0 there exist w E ba(A) and k > 0 such that -k{t :s:: w :s:: k{t and

wll :s:: £,

(iii) for every

£

> 0 there exists an A-simple function such that 1111 - fd{tll :s::

£.

Proof: (i)=>(ii): By decomposing II into 11+ - 11- we may assume that II is nonII(S)/6. negative. There exists 6> 0 such that v(E) < £ when {t(E) < 6. Set k By considering the cases where {t(E) < 6 and {t(E) ~ 6, it is easily checked that v(E) - k{t(E) <

£

(3.5)

for all E E A.

From Lemma 1, 1I /\

k{t(S) == inf{v(EC)

+ k{t(E) : E

E A}

so 1111

11/\ k{tll

II(S) -

11/\

k{t(S) == sup{lI(E)

k{t(E): E E A} :s:: £

by (5). Setting w = II t, k{t gives (ii). (ii)=>(iii): Choose w to satisfy (ii) and then choose f as in (2) so Ilw (Proposition 2.2.1.7(v)). Then 111I - /d{tli :s:: 111I - wll

+ IIw -

fdp.ll:S:: 2£'

/d,..11 :s:: £ + 2£'

and (iii) holds. (iii)=>(i): Given £ > 0 choose f as in (iii). Then there exists 6 > 0 such that lIE /d,..1 < £ whenever {t(E) < 6 [Remark 3.2.2]. Thus, if {t(E) < 6, 11I(E)1 :s:: £ + IIEfd,..1 < 2£ from Proposition 2.2.1.7. We now derive the Radon-Nikodym Theorem for countably additive measures on ".-algebras from Bochner's result. This result follows easily from the approximation property in (iii) and the completeness of LI ({t) with respect to mean convergence. Let L: be a a-algebra of subsets of S. If,.. is a measure on L: and f : S - t R* has a ,..-integral, we denote the set function E - t IE fd{t by fd,... Note that we then have II/d{tll [Exer. 3.2.26].

1/d,..1 (S)

=

is If d,.. = Ilflll I

CHAPTER 3. INTEGRATION

140

Theorem 4 (Radon-Nikodym) Let p. be a a-finite measure on L: and v a a-finite signed measure on L: with v «p.. Then there exists a L:-measurable Junction J : S -4 R such that v( E) = JE J dli Jor all EEL: [i. e., v = fdll}; J is unique up to p-a. e. Proof: By the Jordan decomposition we may assume that v is a measure. We first assume that both p. and v are finite measures. Then v is It-Continuous by Proposition 3.12.5. By Theorem 3 for each kEN there exists a simple function JI; such that Ilv- h:dpil < 1/ k. Since Ilfl;dp hdlill = IIIk - hilI' Ui;} is a Cauchy sequence in Ll(p.) and, therefore, converges to some J E LI(p.) by the Riesz-Fischer Theorem [3.5.1]. Then

implies that v = Jdp.. The extension of the result to the a-finite case is given in the proof of Theorem 3.12.8. Note that the proof of the Radon-Nikodym Theorem above depends on Bochner's version of the theorem for finitely additive set, fuuctions and the completeness of L I (p.) with respect to mean convergence when Ii is a cOllntably additive measure defined on a a-algebra. This proof of Bochner's Theorem basically depends on order properties of the space ba( A) and can be regarded as elementary.

Exercise 1. Let p, v E ba(A). Show Ii V

veAl =

sup{p(E)

+ v(A\E)

:E

c

A, E E A}

defines an element of ba(A) which is the supremum of Ii and v in the order of ba(A).

Exercise 2. Let A be the algebra of subsets of N which are eit,her finite or have finite complements. Define v, Ii on A by

if E is finite and

1+

L 1/21<] kEE

if E is infinite. Show v « p. but there is no p-integrable c: > 0, find a simple function J satisfying (iii).

J

with v

Jdp.

Given

3.13. LEBESGUE DECOMPOSITION

141

Lebesgue Decomposition

3.13

Intuitively, a measure v is absolutely continuous with respect to a measure JL if v is small on sets with small JL-measure (Proposition 3.12.S). \Ve now consider a concept in opposition to absolute continuity, called singularity, which was briefly considered in §2.2.2. Let 2: be a a-algebra of subsets of S. Recall (2.2.2.6) that two measures JL and v on 2: are said to be singular if there exist A, B E 2: such that An B = 0, AU B S, JL(A) = 0 = v(B). If JL and v are singular, we write Two signed measures JL, v on 2: arc singular, written again lilv, if IJLlllvl. If v is a finite signed measure, recall that v+ and V- are singular [2.2.2. To illustrate that absolute continuity and singularity arc opposites, we have Proposition 1 Let

IL,

Proof: Since Ivl Ivl (B)_ But, Ivl

<

()

v be signed measures on 2:. If JLlv and v

<

JL. then v

O.

Ipl, there exist A, BEL AnB 0, AuB = S with 1J1,1 (A) = IJLI implies Ivl (A) = O. Hence, Ivl (S) = Ivl (A) + 1111 (Ii) = O.

We are now going \'0 show that given a J1" any a-finite signed measure can be decomposed uniquely into two parts, one of which is absolutely continuous with respect to IL and another which is singular with respect to JL. This is called the Lebesgue Decomposition. For this theorem, we need a preliminary result. Proposition 2 Let

Vk.

JL be measures an

2:. If Vk lit for every k and

11

then vlJ1,. Proof: Recall that v is a measure [Exer. 2.2.13]. For each k there exist A k ,

Ih

E

2:, Ak n Bk

13 = N

= = n

0,

Ak U Ih = S with JL(Ad = IIk(Bk )

o.

Put, A

A k,

B k . Then

k=l

It(A) ~

I: JL(Ak)

()

k=!

and 00

v(B)

= I: vk(B) k=l

~

I: vk(Bk) =

O.

k=!

Theorem 3 (Lebesgue Decomposition) Let v be a a-finite signed measure on 2: and JL a measure on 2:. Then there exist signed mea8ures Va and Vs on 2: 81!ch that v Va + v. and Va « IL, V.J_/L The decomposition is unique.

CHAPTER 3. INTEGRATION

142

Proof: First assume that v is a finite measure. Let.lVi {E E and set a = sup{ v( E) : E E ./Vi}. Since v 2: 0 and v is finite, 0 ::; a

{Ed C M such that limv(Ek )

= a.

Set E

UE

.=1

k.

2.: : J1( E) <

00.

O} Choose

Then E E M and vee)

[clearly vee) a since E E M, but vee) 2: v(E.) for all k implies vee) 2: We claim that v(A\E) 0 for all A E M.

=a

aJ. (3.1 )

v(A \E) + v( E) > a. Since E U A E .M, this would For otherwise, veE U A) contradict the definition of a. Define v.(A) = v(A\E), v,(A) = v(A n E) for A E 2.:. Clearly v Va + VS' (1) implies that Va «i:: J1. Now J1(E) 0 since E EM and v.(S\E) = v((S\E) n E) = 0 so J1.Lv•. Next assume that v is a t7-finite measure. Let S

UE., where Ek E

k=)

pairwise disjoint and V(Ek)

<

00.

f

Set vk(R)

v(EnE.) for E E

Then v

=

f

(Vk)a and v, (Vk) •. Then Va and v., are measures .=) k=1 from Exer. 2.2.13. Clearly Va ~ J1 and v• .LJ1 by Proposition 2. If v is a t7-finite signed measure, then v v+ V-. In this case sct Va = (v+)a - (v-)a and v. (v+). - (v-)., and apply Exercises 1 and 3.12.1. For the uniqueness, suppose v Va + v. = v~ + v~, where v~ ~ J1 and v~.LIl. If F E 2.: is such that Ivl (F) < 00, t.hen Va V: = V; - v, is both si1lgular wit.h {/<; n F : J~' E 2.:} respect to J1 and absolutely continuous with respect to J1 OIl so Va -v~ V; -v. = 0 on by Proposition 1. Since V is t7-finit.c, Va -v~ = V.: -V, co. 0 From the first part, set Va

on

2.:.

The Lebesgue Decomposition is often derived from the Radon-Nikodym Theorem [[Roy] p. 278J. This usually requires the assumption that both V and It a.re t7finite. Note in the proof above it is only necessary to assume that J1 is non-negati vc, monotone, countably subadditive and vanishes at 0, i.e., J1 is an "outer measure". The proof given here is due to Brooks ([BrD. The t7-finiteness condition in the Lebesgue Decomposition cannot be dropped. Example 4 Let 5 = [0,1] and V be counting measure restricted to the class of Lebesgue measurable subsets of S. Then there is no Lebesgue Decomposit.ion for V with respect to m. For if V = Va + v. as in Theorem 3, there exist measurable A, Jl such that An B = 0, S AU Band v,(A) m(B) = O. Take x E A. Then

v({x}) since m( {x} )

l=va ({x})+v,({x})=O

°

and x E A.

There is a version of the Lebesgue Decomposition Theorem for bounded finitely additive set functions; see [RR] 6.2.4 for this and further such results.

3.13. LEBESGUE DECOA1POSITION

143

Exercise 1. Let VI? V2 be £T-finite signed measures on L with one being finite. Let /l be a signed measure on L with v,J..p,. Show (avI bV2)J../l. Exercise 2. Let E be the even integers. Let v be counting measure and define /l on P(N) by p,(A) = 1/2k . Describe the Lebesgue decomposition of v with respect to /l. Exercise 3. Let f (I.) e t for t :s: 1 and f (t) t ~ 0 and g(t) = 0 for t < O. Let v = fdm, I}' Decomposition for v with respect to /l. Exercise 4. a x

/3

0 for t > 1, 9 (t) (1:1)2 for = gdm. Describe the Lebesgue

Show under appropriate hypotheses that a 1.. /l or

/3

1.. v implies

.i/l x v.

Exercise 5. If a = an + a. (/3 = with respect to /l (v), show (a x

!3.

In.

(a x (3).

+ f3.)

is the Lebesgue Decomposition of a (19)

aa x

a. x f3.

+ a.

x /3a

+ aa x f3.

gives the Lebesgue Decomposition of ax ,8 with respect to /l x v.

144

CHAPTER 3. INTEGRATION

The Vitali-Hahn-Saks Theorem

3.14

In this section we prove a theorem closely related to t.he Nikodym Convergence Theorem, the Vit,ali-Hahn-Saks Theorem. Our proof ofthis result depends on the Nikodym Convergence Theorem, but sometimes the Vitali-Hahn-Saks Theorem is established first and then the Nikodym Convergence Theorem is derived as a consequence [see [DS] for example]. Let 2::; be a a-algebra of subsets of S. Let /l be a measure on 2::; and {Vi} a sequcnce of signed measures on 2::;. Then {v,} is uniformly p.-continuous if lim vl(E) = 0 I'(R)~O

uniformly in i. Theorem 1 Let Vi be a finite signed mcasur'e on 2::; and It a measure on 2::; such thai each Vi is /l-continu01ls. If {VI} is uniformly cOl1ntably additive, then {VI} is unifoT7nly p.-continu01ls. Proof: If the conclusion fails, there exists t > 0 such that for every 6 > 0 there exist k, E E 2::; with IVk(E)1 Land p.(B) < 6. In particular, there exist. HI nl such that Ivn ,(EI )I2: t and p.(El) < 1. There exists 61 > 0 such that. Ivn , (E)I < (/'2 whenever p.(E) < 61 • There exist Ez E nz > 711 such that IV"2(Io'2)1 2: ( and p. (E2) < 61 /2. Cont.inuing this construction produces scqucnces {E k } C 2::;, 6kt I < 6k/2, nk such that Ivnk(Ek)1 2: f, p.(Ek+ 1 ) < IJ k/2 and Ivn.(E)1 < <./2 whellcver It( E) < 15k- Note that

r

p. (

U E1) :::: J=k+!

L

p.(Ej ) < Dk/2

+ 8kt d2 + ... < 8k/2 + 8k/2 2 + ...

8k

j=k+1

so that

Now set Ak = Ek \

U E)'

The {Ad are pairwise disjoint and

J=k+1

by the observation above. However, by t.he uniform conntable addit.ivity of {v,}, we have lim vi(A k ) 0 uniformly for i E N [Lemma 2.8.4]' and we have the desired k

contradiction. Remark 2 Note that only the facts that It is non-negative, increasing and countably subadditive were used.

3.14.

145

THE VITALI-HAHN-SAKS THEOREM

From the Nikodym Convergence Theorem and Theorem 1 we can now obtain the Vitali-Hahn-Saks Theorem. Theorem 3 Let Vi be a finite signed measure on Land Ji a measure on L such that each Vi is Ji-continuous. Iflimvi(E) = v(E) exists for each EEL, then

.

(i) {v;} is uniformly Ji-continuous and (ii) v is Ji-continuous. Proof: By the Nikodym Convergence Theorem {Vi} is uniformly countably additive so (i) follows from Theorem 1. (ii) is immediate from (i). Both the Vitali-Hahn-Saks and Nikodym Theorems fail for countably additive set functions defined on algebras [see Exercises 1 and 2]. There are versions of the theorems for certain finitely additive set functions [see [DUll. Exercise 1. Let A be the algebra of subsets of [0, 1) generated by intervals of the form [a, b), 0 ~ a < b ~ 1. Define Vk on A by vk(A) = 2km([1 - 1/2k, 1) n A) so Vk is countably additive on A. Show that if b < 1, vk[a, b) = 0 for large k and Vk[C, 1) = 1 for large k so limvk(A) = v(A) exists and v(A) = 1 if [1 - 8,1) c A for some 8 > 0 and v(A) = 0 otherwise. Show v is purely finitely additive (§2.6.1) so (ii) of Theorem 2.8.5 fails. Show also that (i) of Theorem 2.8.5 and (i), (ii) of Theorem 3 fail. Exercise 2. Let A be the algebra of subsets of N which are either finite or have finite complements. Define Vn on A by Vn (A) = n if A is finite and n E A, Vn (A) = -n if N is finite and n E AC and Vn (A) = 0 otherwise. Show each Vn is bounded, finitely additive and lim Vn (A) exists for each A E A. Let Ji (A) = L 1/2n when A is finite and Ji (A) = 1

Vn

«

+

nEA

L 1/2 n when AC is finite. Show Ji is bounded, finitely additive, nEA

Ji but (i) fails.

Chapter 4 Differentiation and Integration 4.1

Differentiating Indefinite Integrals

We consider the first half of the Fundamental Theorem of Calculus (FTC), the differentiation of indefinite integrals. We consider indefinite integrals in Rn. Points x in Rn are denoted by x = (XI, .•. ,xn ) where Xi E R. If X E Rn and r > 0, the open cube with center at x and sidelength 2r is denoted by

8(x,r)

{y:IXi-Yil
1, ... ,n}.

All statements concerning measurability, integrability and almost everywhere refer to Lebesgue measure on Rn which is denoted by m. We begin by establishing a covering theorem which will be used in the proof of the differentiation results. Lemma 1 Let C be any collection of open cubes in Rn and let U = U{ I : I E C}. If k

C

< m( U), there exist pairwise disjoint

S1, ... ,8k E C such that

L

m( Sj) > 3- n c.

)=1

Proof: By regularity of m, there exists a compact K C U such that m( K) > c so there exist finitely many cubes II"", I j E C covering K. Let S1 be the {Ii} with the largest sidelength. Let S2 be the {I;} with the largest sidelength which is disjoint from SI and continue this procedure until the {Ii} are exhausted. This gives a pairwise disjoint sequence 81, ... , Sk E C. Let SI be the cube with the same center as 8 i but with sidelength three times that of Si. If Ii is not one of the SI, . .. , there exists Sl such that Ii n Sl f. 0 and the sidelength of Ii is at mosi that of Se. k

Hence, Ii C S;. Then

f{

C

U S; so £=1 k

C

< m(K) S

L

m(8;)

£=1

A function f : Rn -> R is said to be locally integrable if f is Lebesgue measurable and is m-integrable over every compact subset of Rn. The class of all locally integrable functions on Rn is denoted by LfoARfi). '\ 147

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

148

Let f E Lloc(Rn). We consider the problem of differentiating the indefinite integral of f, IE fdm. That is, for what x does the limit, lim r-O

r

is(x,r)

fdrn/rn(S(x,r»,

exist and for what x does the limit equal the integrand, f(x)? Note that for n = 1, we are asking that x+r lim fdm/2r f(x); T-.O+

1

x-r

we will see that information about the usual derivative, lim

h-tO

x +h

1 x

fdm/h

= f(x),

can be derived from our general results. The principal tool used in our proof is the Hardy-Littlewood maximal funct.ion. The Hardy-Littlewood Maximal Theorem gives bounds on the average vallie of a function and this result is used in the differentiation theorem. If f E Lloc(Rn), x ERn and r > 0, we set

r

,1 r f(x)

is(x,r)

fdm/m(S(x, r)),

the average value of f over S(x,r), and Mf(x)

supAr If I (x). Mf is called the r>O

Hardy-Littlewood maximal function. Proposition 2 M fis measurable. [In fact, M f is lower semi-continuous.} Proof: We show E = {x : M f( x) 'S a} is closed for every a > O. Suppose Xk E E and Xk --; x. Let Ak be the symmetric difference S(Xk' r)ll.S(x, r) and fk CAJ. Since

nAk

0, fk

--t

0 and Ifkl 'S If I so by the OCT

JRn

Ifkl dm

--t

O. Since

k=l

S(x,r) C Ak u S(Xk' r),

,1r If I (x) ~ IAk If Idm/m(S(x, r») + IS(""r) If Idm/m(S(xk' r)) ~

IRn

Ifkl dm/m(S(x, r))

Hence, AT If I (x) ~ a and Mf(x) ~

aj

+ a.

i.e., x E E.

Theorem 3 (Maximal Theorem) There exists a constant C > 0, depending only on n, such that f01' all fELl and a> 0, mix : Mf(x) > a} 'S (C/a) JRn If I dm. Proof: Let E. {x: Af f( x) > a}. For each x E E. choose r x > 0 such thaL AT. If I (x) > a. The cubes {S(x, rx) : x E Ea} cover Ea so if c < m(Ea), by Lemma 1 there exist Xl, ... ,Xk E E. such that the cubes {S) S(x),rx,): j = I, ... ,l:} a,re pairwise disjoint and

Since c

< m(E.)

is arbitrary, the desired inequality follows.

4.1. DIFFERENTIATING INDEFINITE INTEGRALS Theorem 4 If f E LloARn), then lim Arf(x) r-O

= f(x)

149

for m-almost all x ERn.

Proof: It suffices to show Arf(x) f(x) for almost all x in the cube S(O,N) for arbitrary N. For x in the cube and l' < 1, the values of Arf only depend on the values of fin S(O,N + 1) so we may assume that f E Ll(Rn). Let c> 0 and pick a continuous function with compact support 9 on Rn such that Ian If gl dm < t' [3.5.6]. Since 9 is uniformly continuous, for every x E Rn and 5 > 0 there exists l' > 0 such that Ig(y) - g(x)1 < 5 when y E S(x,1') so

IArg(x) - g(x)1 = Hence, lim Arg(x) r_O

= g(x) f(x)1

g(x))dyl /m(S(x,1')) < 5.

and

= r_O lim IAr(J ~

Let En

IflS(x,r) (g(y)

AI(J

g)(x) g)(x)

+ (Arg(x) -

+ Ig

+ (g(x)

f(x))1

fl (x). {x: If - gl (x) > a}. Then gl dm 2: am(Fa) so by the

{x: lim IATf(x) - f(x)1 > a} and Fa r-a

En C Fa/2 U {x : M(J - g)(x) > a/2}. But ( >

g(x))

IFa If -

Maximal Theorem,

g)(x) > a/2}

~

21: a

+

2C a

Hence, m(Ea) 0 for all a > 0, and the result follows. We now show that Theorem 4 can be improved by moving the absolute value in the conclusion, lim

If

T-O lS(x,r)

fdm/m(S(x,r)) - f(X)1

= r-O lim 1f (J lS(x,r)

f(x))dm/m(S(x,

1'))1

0,

inside the integral sign. A point x is called a Lebesgue point of f if lim

f

T-als(x,r)

If - f(x)ldm/rn(S(x,r))

=0

and the collection of all Lebesgue points of f is called the Lebesgue set of f. Theorem 5 If f E Ltoc(R n), then almost every point of Rn is a Lebesgue point of f. Proof: By Theorem 4, for every q E Q there exists an rn-null set Zq such that lim

f

r-als(x,T) for x E Z~. Set Z

U Zq.

If(y) - ql dy/rn(S(x, r)) Then rn(Z)

qEQ

P E Q with If(x)

pi < (;..

o and if x

= If(x) - ql rf- Z for any

Since

If(y) - f(x)1 ~ If(y)

pi + e,

E

> 0, there exists

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

150

r

lim If(y) T~O is(er,,) so lim

r

,~O is(x,,)

f(x)/dy/m(S(x,r»::::lf(x)-pl+t<2E

If(y) - f(x)1 dy/m(S(x,r)) = 0 for x

1c z.

Next we show that the limit in Theorem 5 above exists when more genera.l sets than cubes are considered. These results are particularly useful in R where lltey yield differentiation results.

Definition 6 Lei x ERn. A family of Borel sets, {E, : r > OJ, shrinks regularly 10 x if E, c S( x, r) for every r > 0 and there exists a > 0 (independent of r) such that m( E,) :::: am( Se x, r» {x need not belong to E T }.

Example 7 In R, the families ([x -r, x) . r > OJ, {[x, x+r) : r > O} shrink regularly to x [with a = 2]. In Rn, if E, {y: IIx yll < r}, then {E shrinks rcguliuly to x. T }

For an example of a family of rectangles in R2 whose intersectioll is {x} but which do not. shrink regularly t.o x see Exercise 1.

Theorem 8 Let f E LlocCRn). If x is in the Lebesgue set of f and shrinks regularly to x I then lim

1'_0

m

(1",) /'.;1'

r

jEr

{I~T

: r > O}

If(y)-f(x)ldy=O.

Hence, lim

1'-0

rn·

(IE) T

r

jBr

f(y)dy = f(x).

In particular, this holds for almost all x ERn.

Proof: Let a ~s in Definition 6. Then ;:;,(k,)JE,lf(y)-f(x)ldy::::

Js(x.T)lf(y)-fex)ldy (1.1 ) ~T----rrJS(X.T) If(y)

f(x)1 dy

-+

0

by Theorem 5. The last two statements follow from (1) and Theorem 5. From Theorem 8 and Example 7 we obtain a version of the FTC in R. Let f : [a, b]-+ R be integrable and let F(x) = J: fdm be the indefinit.e integra.l of f.

Theorem 9 (FTC) Fis differentiable at every Lebesgue point x of f with F'(x) fex). In particular, this holds a.e. in [a,b].

=

151

4.1. DIFFERENTIATING INDEFINITE INTEGRALS

Finally, we consider the differentiation of arbitrary regular Borel measures on Rn [The regularity assumption is redundant (2.7.6); we make this assumption for the reader who has skipped §2. 7. J Let v be a regular Borel measure on R n. We say that II is differentiable at x if lim II(S(X, r))fm(S(x, r)) exists; the value of this limit r....O is denoted by Dv(x) and is called the derivative of II at x. If v is an indefinite integral of a locally integrable function, this agrees with our previous definition of the derivative. Lemma 10 If II is a regular Borel measure on Rn and v(A) for almost all x E A. { x E A : ~i.To;,\~\;~)l)

Proof: For 6 > 0 let As

0, then Dv(x) = 0

> 6}. It suffices to show that

m(As) = 0 for every such 6. Let f > O. By regularity there is an open V :J A such that v(V) < (. For each x E As there is an open cube with center at x, S", C V, such that II(S,,) > m(S",)6. By Lemma 1 if U U Sx and c < m(U), there exist xEA.

xl.' .. ,Xk

E As such that SXl' ...

,Sx~

are pairwise disjoint with

k

C

< 3n L.: m(S.,;) 1=1

:::: (3 n /6)v(V) < (3 n j6)c Therefore, m(U) :::: (3"/6)f and since As C U and f > 0 is arbitrary, m(As) A signed Borel measure II is said to be regular if jill is regular.

O.

Theorem 11 Let v be a regular signed Borel measure on Rn and v Va + Vs be its Lebesgue Decomposition with respect to m with f = ~, the Radon-Nikodym derivative of Va with respect to m. Then v is differentiable a.e. with Dv = f a.e. Furthermore, for almost all x, limll(Er)/m(Er ) = f(x) whenever the family {Er : r > O} shrinks r-'O regularly to x. Proof: By the Jordan Decomposition we may assume that v is a measure. The first part of the Theorem follows from Theorem 4 and Lemma 10. The last statement follows from Theorem 8 and Exercise 2. Exercise 1. Let Er = [-r,rJ x [_r 2 ,r2 ]. For r:::: 1, show Er C S(O,r), nr>oEr = {(O,O)}, but {Er: r > O} doesn't shrink regularly to (0,0). Exercise 2. Let v be a Borel measure on Rn and v(A) = O. Show that for almost all x E A, limv(Er)jm(Er ) = 0 when {Er: r > O} shrinks regularly to x [Lemma IOJ. r~O

Exercise 3. Show Arf(x) is continuous in Exercise 4. Let f : [a, b] at x, show F'(x) = f(x).

-+

T.

R be integrable and F( x)

= J: f dm.

If f is continuous

152

CHAPTER 4. DIFFERENTIATION AND INTEGRATiON

4.2

Differentiation of Monotone Functions

We use the results of the previous section to establish a result of Lebesgue on the almost everywhere differentiability of a monotone function. Again all measurability statements refer to Lebesgue measure. Let f : R ---+ R be monotone [assume for definiteness that f 1]. Then f is not continuous at x if and only if f(x+) - f(x-) > 0, i.e., if and only if f has a jump discontinuity at x. If x is a discontinuity of f, we may choose a rational T'x satisfying f(x-) < rx < f(x+) and obtain a 1-1 mapping x ---+ T';c from the set of discolltinuiLies of f into Q. This shows that thc set of points of discontinuity of a monotone function is countable. Hence, a monotone function has points of continuity in evcry open interval We first consider the case of left continuous functions. Such functions induce Lebesgue·Sticltjcs measures and Theorem 4.1.11 can be used to prove their a.e. differentiability. Theorem 1 Let f with fl DJlj a.e.

:R

---+

R be

1 and

left continuous. Then

f is diJJcr-nttiablc a.c.

Proof: Let III be thc LebesgueSticltjes mcasure induced by f· Then Dllj exists a.e. [4.1.11]. The families {[x h,x): h > O} and ([x,x+h): h > O} shrink to x [Example 4.1. 7] so the limits

lim Jl![x

h~O+

h, x)/h

lim

h~O+

and

exist for almost all x by Theorem 4.1.11. Thus, f'(x) exists for almost all x and f' DJl J a.e. We now remove the left continuity assumption from Theorem L Let f : R ---+ R be 1- Define f. : R -> R by f.(x) = J(x-) = lim J(y). y-x-

Then f. :::; J, f. T and f. is left continuous [for each x there exists Xk T x such that f is continuous at Xk so f(x-) = limJ(xd J.(x) limf.(xk)]. Similarly, f. (x+) f. (x) = f(x+) f(x-) so f and f. have the sa.me points of continuity. Lemma 2 If f. is (iifferent.iable at x! then f is differentiable at x with f; (x) Proof: Let m f.

=

J' (x).

f;(x). Note f(x) = J.(x) since f. is continuous at x. Let f.(x))/(y - x) < m + e when

> O. Choose 6 > 0 such that m - ( < (f.(y)

4.2. DIFFERENTIATION OF MONOTONE FUNCTIONS

o < IY - xl < 6. Yk

1 y with 0 <

Fix Y with 0 < Iy - xl < 6 and choose a sequence {yd such that xl < 6 and f continuous at each Yk. Then

IYk

m ~

153

t

< (J.(y)

f(x))/(y - x) ~ (J(y) - f(x))/(y - x) lim(J.(Yk) - f.(X»)/(Yk - x) ~ m

(J(y+) - f(x))/(y - x)

+t

so

1(J(y) when 0

< Iy - xl < 6. Hence, f'(x)

Theorem 3 (Lebesgue) If f: R

f(x))/(y

x) -

ml ~

t

= m.

-4

R is 1, then f' exists a.c.

Proof: Lemma 2 and Theorem 1. There are proofs of Lebesgue's Theorem which do not use measure theory. See [RN] L2 for a proof due to F. Riesz. This result cannot be improved to read that f' exists except for a countable set of points. Indeed, we have Proposition 4 If E C [a, b] has measure 0, there is a continuous increasing fun~tion f : [a, b] ~t R such that f'(x) = 00 for every x E E. Proof: For each kEN choose a bounded open set G k => E with m(G k ) < 1/2k. Set fk(X) = m(G k n [a, x]). Then fk is continuous, increasing and fk ~ 1/2k on [a,b].

If f

f

fk' then f is continuous, non-negative and increasing. If x E E and h > 0

k=!

is sufficiently small, the interval [x, x

m(x,x + h] h. Thus, (fk(X h > 0 is sufficiently small,

+ h)

+ h]lies in G k so for snch f(x»/h

h, fk(x + h) - fk(X) =

1 for small h > O. For every N if

N

(f(x + h) - f(x»/h :2: "L,(fk(X + h) - fk(X»/h

= N.

k=!

Hence, f'(x)

00.

Notation: If I is an interval in Rand f : 1--> R is differentiable a.e. in I, we denote by j the function defined on I by j(t) f'(t) if f'(t) exists and j(t) = 0 otherwise. Concerning the integrability of the "derivative" of a monotone function, we have •

b •

Theorem 5 If f ; [a, b] -. R is increasing, then f is integrable and fa f f(a). I

~

f( b) -

154

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

Proof: Extend f to [b, (0) by setting f{t) feb) for t b. For each k set (J(t + l/k) f(t))/(l/k) for a ~t ~ b. Note gk -> a.e. and gk ::::: O. By Fatou's Theorem,

gk(t)

f:

gk dm

= limU: kf(t + l/k)dt -

· (k l1m

lk fb+I a+Ilk

fd m )k ,fab fdm)

lim(J(b) - k f:+I lk fdm)

f: kf(t)dt)

= ll'm(k JIfbb+l/k f'lm

= feb)

<.

k J~"+I/k fdm)

-limk f:+ 1 / k fdm ~ feb)

f(a).

Strict inequality in Theorem 5 can hold.

Example 6 Let K be the Cantor set in [0,1] and let [0,1]\1< =

= U

1;: as ill

n-;::;:::O

Example 2.5.7. Define f(t) for t E 1;: [make a sketch]. Then f is increasing and continuous on J{c and the range of f is dense in [0,1] so f can be extended to a continuous function, f, on [0,1] [f(x) inf{f(t): t ~ x, t E j{C}]. Obviously, f'(t) = 0 for t EKe so f' 0 a.e. But, f~ 0 < fell f(O) = J.

i

The function f constructed above is called the Cantor function.

Remark 7 Increasing functions f with the property that f' 0 a.e. arc called singular fllnctions. There are examples of strictly increasing singular functions; see [Fre], [RN] p. 44-49, or [HS] p. 278-282. For two entertaining articles on singular functions see [Cat] and [Za]. We address the question of when equality holds in Theorem 5 of section 4J1. We next present an interesting result on the termwise differentiation of a series due to Fubini.

Lemma 8 Let fk : [a, b]

10.

Then

-+

R be increasing and assume

f ik converges a.e.

k=l

Proof: Since 10 is also increasing, II. exists a.e. for all k ::::: O. ][ E {t : IH t) exists for all k ::::: OJ, then m(EC) = 0 and 0 ~ Ik(t) < 00 for tEE, k ::::: O. Let Sn

=

t ik. Since (fo

sn)

k=l

on E [since s~+l

::::: s~l.

r for n ::::: 1, I~ : : : s~ on E which implies that lim s;, exist s n

That is,

ik converges a.e.

Theorem 9 (Fubini) Let!k; [a, b] pointwise to

10.

Then

io =

->

fI" a.e.

R be increasing and assume

4.2. DIFFERENTIATION OF MONOTONE FUNCTIONS

Proof: The series

155

jl< converges a.e. by Lemma 8. By replacing fk by fk - fk( a)

if necessary, we may assume that fl< 2:: O. It suffices to show that there exists a subsequence {s".} of

Ii

51<

such that 8".

--t

a.e. since 8n ~ 8n +1. For each

j

k there exists nk with 0 ~ fo(b) 5 n .(b) < 1/2 k where we may assume nk < nk+I' Since (fo s".) for each t E [a, b],

r,

'~nk) converges uniformly on

Therefore, the series

[a, bl. Since (fo - 5".)

.~".) converges a.e. so 8nk

follows from Lemma 8 that

--t

j

r, it

a.e.

Dini Derivates: In the calculus we learn that a function f : [a, b] --t R which has a non-negative derivative in [a, bl is [this follows from the Mean Value Theorem]. We consider a strengthening of this result. The material which follows is used only in section 4.3 and can be skipped by the reader who does not wish to go through this section. Let I [a, bl and f : I R. The Dini derivates of f at x E I are defined to be d+f(x)

f(x))/{t - f(x))/(t

d+f(x) d-f(x) d_f(x)

x), x),

- f(x))/{t - x), (f(t) - f(x))/(t - x).

The upper (lower) derivative of f at x is defined to be (I2J(x)

limt_x(f(t)

f(x))/(t - x)).

[See Exercise 4 for the definitions.] Thus, f is differentiable at x if and only if all four Dini derivates of f at x are equal and finite if and only if the upper and lower derivatives are equal and finite. The derivative is then the common value. Some of the elementary properties of derivates are given in Exercise 6.

Proposition 10 Let f : I R be continuous. If c > (f(b) - f(a))/(b - a), then at uncountably many points x of (a,b) we have d+f(x) ~ c. [SimilaT"iy, if (f(b) f(a))/(b - a) > c, d- f(x} 2:: c.] Proof: Set k

c( b

a) and consider the function

g(x)

f(x)

f(a)

k(x

a)/(b

a).

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

156

= 0 and

Then g(a)

g(b)=f(b)

f(a)

k
Let s be such that 0 = g(a) > s > g(b). Consider the set E {t E [a,b]: get) 2: s}. Since 9 is continuous on [a, b], this set is closed, and, therefore, compact. Hence, if t. supE, by continuityofg, g(t.) = s, and sinceg(b) < s, t. < b. Sinceg(ts+h) < s for sufficiently small h > 0, d+g(t.) SO so

d+ J(t.) = d+g(t.)

+ kJ(b

a) S kJ(b

a)

c

[Exercise 6]. Different s's generate different t.'s and there are uncountable many s's between 0 and g(b) so there are uncountably many points t E (a, b) such that d+ J(t) S c. This result is a substitute for the :VIean Value Theorem for non-differentiable functions.

Theorem 11 IJ J is continuous and one Dini derivate is non-negative except perhaps for a countable number' oj points in [a, b], then J is inc1·easing. Proof: Suppose d+ J(t) 2: 0 except possibly for count ably many points t in [a, b] [d+J(t) 2: 0 implies this and the case d- J(t) 2: 0 is similar]. If J is not increasing, there exist x < y such that J(x) > fey). Use Proposition 10 with

(f(y) - J(x)J(y

x) < c < 0

on [x, yj to obtain that d+ J(t) < 0 at uncountably many points in [x, yj. This contradiction establishes the result. This result obviously covers the usual calculus test for llH.H:;d.Dlll1'. functions. More information on Dini derivates can be found in [Boa].

Exercise 1. Show that if

J:R

-->

R is monotone,

J is

measurable.

Exercise 2. Let D {td be any countable subset of R. Show there exists an function J whose discontinuities are exactly D. [Hint: Choose ak > 0 such ak

<

00.

Define !k(t) = 0 if t < tk and !k(t)

= ak

if t 2: tk. Put J

f

Jk

k=!

J is continuous in DC and has jump ak at tk']

Exercise 3. Show that a continuous, nowhere differentiable function is not monotone on any non-degenerate interval. Exercise 4. Let is defined by

J; [a, b]

I

-->

R and x E I. The limit superior of

limJ(t) = infsup{J(t): t E 1,0 < i--x

£>0

It - xl < o:}.

J at x,

(t),

4.2.

DIFFERENTIATION OF MONOTONE FUNCTIONS

157

Define limit inferior, lim I (t), and establish the analogues of the statements in Exert_r

cises 1.2.1 and 1.2.10. Define one-sided such limits.

Exercise 5. Let I(t) = tsin(l/t) if t -# 0 and 1(0) O. Compute the four Dini derivates of I at O. Now construct a function all of whose Dini derivates are not equal at O. Exercise 6. Show !lJ(x)+dg(x) ::; 4(1 + g)(x) (dl(x) +dg(x) ;::: d(l +g)(x)), where 4 (d) is any lower (upper) derivate. Show that if f'(x) exists, then d(l + g)(x) = f'(x) + dg(x), 4(1 + g)(x) f'(x) + dg(x). Exercise T. If I

i,

show any derivate is ;::: O.

Exercise 8. If I assumes a local maximum at x, show d+ I(x) ::; 0 and d_/(x) ;::: O. Exercise 9. If max{ D I( x), D I(x)}

< 00

,

show I is continuous at x.

Exercise 10. Show Theorem 5 can be improved to

Exercise 11. If I in Theorem 5 is not continuous, show strict inequality holds.

158

4.3

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

Integrating Derivatives

In this section we consider the other half of the Fundamental Theorem of Calculus (FTC), the integration of derivatives. We begin by showing that the most general (and most desirable!) form of the FTC does not hold for the Lebesgue integral by giving an example of a derivative which is not Lebesgue integrable. Again all measurability statements refer to Lebesgue measure. Example 1 Let f( t) t.iable on [0,1] with

t 2 cos( 1r /t 2 ) for 0 < t :::; 1 and f(O)

for 0 < t :::; 1 and 1'(0) = O. For 0 is (Riemann) integrable with

If we set bk

1/V2k,

ak

< a < b < 1, I' is bounded on [a, b] and, therefore,

I:: f'dm

+ 1), then Ii Il'ldm ~

j27(4k

{[ak,b k]} are pairwise disjoint so

= O. Then f is differcIl'

1/(2k). The intervals

I:: 1f'ldm

~

f

1/(2k)

00.

k=1

Hence, f' is not integrable over [0,1]. We establish the most general form of the FTC for the Lebesgue integral. Lemma 2 Let [a,

f : [a,b]

->

R. Suppose all Dini derivates are non-negative a.c. m -00. Then f is inereasing.

b] and that no Dini derivale is

Proof: Let E be the set of points where some derivate is negative. By 4.2,4 there exists an increasing function g : [a, b] -> R such that g'(X) 00 for every x E E. Let, c > 0 and set h = f + Eg. At all points of [a, bj\E all derivat.es of h arc non-nC'gative for 1; E }; [Exercises 4.2.6 and 4.2.7J. The same holds al. points of E since gl( x) = and the derival.es of f at these points are not -00. Thus, by 4.2.11, hi. lIenee, jf x < y, f(x) + tg(x) :::; fey) + tg(y), and letting f -> 0 gives f(x) :::; fey)·

=

Theorem 3 (Fundamental Theorem of Calculus) Let entiable on [a, b]. If f' is Lebesgue integrable over

l

J'dm

= f(b) -

f : [a, b]

b], then

f(a).

-t

R be d~ffer-

4.3. INTEGRATING DERIVATIVES

159

Proof: Definegk(t) f'(t)iff'(t):::: k and gk(t) = kiff'(t) > k. Then 19k1:::: 1f'1 and 9k - t I' pointwise on [a, b]. Set 'k(t) f(t) gkdm.. We claim that 'k T. First, observe that 'k is differentiable a.e. with r~ I' - gk ::::: 0 a.e. [4.1.9]. Since gk :::: k on [a, b], J:+ h 9kdm./ h :::: k for x E [a, b], h sufficiently small. Thus,

J:

('k(x+h)

rk(x))/h

so no derivate of 'k is

f(b)

(f(x+h)

(,,+h J(x))/h- Jx gkdm./h::::: (f(x+h)-J(x))/h-k

-00. By Lemma 2, rk f(a)::::: J: gkdm.. By the DCT, limJ: gkdm.

T.

In particular, rk(b) ::::: rk(a) or J: I'dm so f(b)- f(a) ::::: J: I'dm..

Replacing f by the reverse inequality. Of course, the annoying feature of Theorem 3 is the need for the hypothesis that I' is Lebesgue integrable. [Recall that the Riemann integral suffers this same defect.] Example 1 shows that this assumption is necessary. It would be desirable to have a theory of integration for which the FTC holds in full generality, i.e., a theory for which all derivatives are and the formula in the FTC holds. Two such theories were developed by Perron and Denjoy [see [Pel for a description of these integrals]. Lately, a very integral which is but a slight variant of the Riemann integral, called the gauge integral, has been given independently by Kurzweil and Henstock ([Ku], [He]) for which the FTC holds in full generality [see [DeS] or [Me] for a description of the integral]. The statement in Theorem 3 cannot be improved to "I' exists a.e.". See the Cantor function in 4.2.6. However, the statement can be improved to "I' exists except for a countable number of [Co] 6.3.10, [liS], p. 299, [Wall.

160

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

Absolutely Continuous Functions

4.4

In this section we consider a form of the FTC in which the functions are only differentiable a.e. [again all measurability statements refer to Lebesgue measure]. In particular, for increasing functions we are asking when equality holds for I.be inequality in Theorem 4.2.5. Recall that if I : [a, b]---> R is different.iable a.e., we denote by j the function jet) f'(t) when f'(t) exists and jet) = 0 otherwise. Suppose I : [a, b] ---> R is increasing and f( x) - f( a) = f: jdm for a <::: x <::: b, i.e., equality in Theorem 4.2.5 for all a <::: x <::: b. Since lim fEjdm = 0 [3.2.17], for m(E)-O

> 0 there exists 6 > 0 such that whenever {(ai, bi)} 7=1 is a pairwise disjoint sequence of subintervals of [a, b] with f: (b i ail < 6, then every

f

i=l

n 1,'b in jdm = ~ If(b.) -

~

I

l(ai)1 <

E.

Functions which satisfy this condition are called absolutely continuous; we give the formal definition.

Definition 1 Let f : [a, b] -, R. Then f is absolutely continuous if for every ( > 0 there exists 6 > such that whenever {(ai, b.)} 7=1 is a pairwise disjoint sequence of

°

open subintervals of [a, b] with

a;) < 6, then If(b;) - f(a,)1 <

L

Example 2 The function f : [a, b] -> R satisfies a Lipschitz condition if there exists L> 0 such that If(x) f(y)1 <::: Llx-yl for x, y E [a,b]. Such a function is obviously absolutely continuous. [The converse is false; see Exercise 7.] For examples of functions which satisfy a Lipschitz condition, see Exercise 1.

Proposition 3 Lei f : [a, b] ....... R be absolutely continuous. Then (i) f is continuous and (ii) f is bounded variation. Proof: (i) is clear. For (ii), let c = 1 and 6 be as in Definition 1. Partition [a, b] by Xo < Xl < ... < Xn b}, where Xi+l - Xi < O. Then Var(f : [x;, Xi+l]) <::: 1 for i 0, ... , n 1. Hence, Var(f : [a, bJ) ::; n. We show below that the converses of (i) and (ii) are false [see Examples 5 and 9].

P

= {a

Corollary 4 If f : [a, b] integrable.

--->

R is absolutely continuous, then f' exists a.e. and

j

is

4.4. ABSOLUTELY CONTINUOUS FUNCTIONS

161

Proof: By Proposition 3 and Theorem 8 of Appendix Al, f is the difference of two increasing functions, so the result follows from Theorem 4.2.3. Example 5 The function f in Example 4.3.1 is uniformly continuous but not lutely continuous by Corollary 4.

abso~

Algebraic properties of absolutely continuous functions are given in Exercises 2-6. \Ve use Lebesgue-Stieltjes measures to describe the relationship between abso~ lutely continuous set functions and absolutely continuous functions on intervals in

R. Let f : [a, b] ..... R be left continuous and have bounded variation. Then f = 9 h, where 9 and h are left continuous and increasing [Appendix AlJ. Extend f (g and h) to R by setting f(t) f(b) [g(t) g(b), h(tl h(b)] for t > band f(t) f(a) [get) g(a), h(t) h(al] for t < a. Then 9 and h are bounded, left continuous and increasing and, therefore, induce finite Lebesgue-Stieltjes measures lig and lih. lIence, lij lig lih is a finite, signed measure; we call lij the Lebesgue-Stieltjes signed measur'c induced by f. Note lij[a,p) = f(p) - f(a) for a:::; {J.

Theorem 6 Ii j

«

m if and only if f is absolutely continuous.

Proof: ¢=: Let A C (a, b) have m-measure O. Let E > 0 and let 5 > 0 be as in Definition 1. There exists a pairwise disjoint sequence of open intervals in (a, b),

{(ai, bi)}, such that

U(ai, b

i)

ail < 5 [2.5.3]. For each n

:> A and feb. i=l

i=l

n

I: 11i/[ai,bi)1 <

If(b;) - f(ai)1

E

i=1

so by Exer. 2.6.6

I~JlJ[ai,bi)! = IliJ(,9(a;,b;))!:::;

E.

Hence, liJ(A) 0 and IliJI (A) = 0 by 2.2.1.7. =>: Note lif[a,{J) liJ(a,{J) for a < (J. Since liJ is finite, for every E > 0 there exists 8 > 0 such that meA) < 8 implies IliJI(A) < t [3.12.5J. Suppose {(ai,b i )}7=1 is a pairwise disjoint sequence from [a,b] with E(b; ;=1

(j

{i: f(b i )

f(a;)

~

O} and

r

ail < 8. Let

{i: feb;)

f(a;) < OJ.

Then

I: If(b.) iEu

and

so

E If(b;)

i::;l

f(ai)1 < 2E.

f(ai)1 <

t

CHAPTER 4. DlFFERENTIATJON AND INTEGRATJON

162

Theorem 7 Let f : [a, b] -; R be differentiable a.e. Then f is absolutely contimwus if and only if j is integrable over [a, b] and f(x) - f(a) jdm for a'::: 3; '::: b.

f:

Proof:

=}:

By Theorem 6 /ll

a.e. Hence, 111 [a, x) = f(x) - f(a) {=: 3.2.17.

«m, and by Theorems 4.1.11

and 4.1.9

f: jdm.

Remark 8 Theorem 7 is sometimes referred to as the FTC for the Lebesgue integra.1Note that Theorem 7 gives necessary and sufficient conditions for equality to hold in Theorem 4.2.5. Example 9 The Cantor function [4.2.6] is continuous, increasing, and hence bounded variation, but is not absolutely continuous by Theorem 7. A function f : [a, b] -; R of bounded variation which is such that f' 0 a.c. is said to be a singular function. The Cantur function supplics all cxaillpic ur a non-constant singular function. Theorem 10 Let f E BV[a, b] be lcft continuous. Then f is sin.gular if and only if /ll .L m. Proof: Let III = (Ill)a + (111). be the Lebesgue Decomposition of Ilf with rcspect to m. By Theorems 4.1.11 and 4.1.9 f' = Dllf a.e. The result now follows. Theorem 11 (Lebesgue Decomposition) Let f BV[a, b]. Then ther'c el:isl9, 9 + h with 9 absolutely continuous and h singular. The h : [a, b] -; R such that f decomposition is unique up to a constant.

f:

Proof: Set g(x) = jdm and h = f - g. Then 9 is absolutely cuntinuuus by Theorem 3.2.17 and h is singular by 4.1.9. Uniqueness follows from Exercise 8. Change of Variable We use the results above to establish a change of variable theorem for the Lebesgue integral. Theorem 12 Let 9 : [a, b] -; R be increasing and absolutely cont£nuOU8 and set a g(a), fJ == g(b). Let f : [a,fJ] -; R* be Lebesgue integrable over [a,j3]. Then

(f 0 9)

9 is Lebesgue

integrable over [a, b] and

fb

Ja

. f(g(t) 9 (t)dt

rfl

= J"

f(x)dx.

( 4.1)

4.4. ABSOLUTELY CONTINUOUS FUNCTIONS Proof: First suppose

163

f is the characteristic function of a half closed interval

b,6) C [a,.8]. Let c = inf{t: get) = I}, d = sup{t : get) J:f dm =6

g(c)=l dgdm

g(d)

I

6}. Then fog

G[c,d)

so

l(fog)ildm

or (1) holds. It follows that (1) holds for S-simple functions, where S is the of all half-closed intervals. Now suppose f is Lebesgue integrable over [a, .81. By Mikusinski's Theorem (3.8.3) there exists a sequence of S-simple functions {J..} such that

Mx)

f(x) for any x for which

f

k=!

We show the sequence {(fk function (f 0 g) il. First,

0

f

k=!

Ifk(x)1 <

00

and

It Ifkl dm <

It fdm =

f

k=!

00,

It fkdm.

g) il} satisfies the conditions of Exercise 3.8.1 for the

by the part above. Suppose 00

\Mg(t»

il

(t)\

< 00.

k=!

If il (l)

0, then

= f(g(t)

9 (tl

fk(g(t» il (t) = f(g(t»

9 (t).

fk(g(t)) il (t) while if 9 (t) > 0, 00

L k=!

By Exercise 3.8.1 (f 0 g)

9 is Lebesgue integrable and

There are other useful conditions under which equation (1) holds. For example, (1) holds if f is bounded and integrable and 9 is absolutely continuous. See [St2j for a thorough discussion of the validity of (1). It should be pointed out that (1) does not hold in general even when f is integrable and 9 is absolutely continuous (Exer. 13). For change of variable in Lebesgue integrals in Rn, see [Ru2J. Exercise 1. Show that if Lipschitz condition.

f : [a, hJ

R has a bounded derivative, then

f satisfies a

CHAPTER 4. DIFFERENTIATION AND INTEGRATION

164

Exercise 2. Let f, 9 : [a, bl R be absolutely continuous. Show If I, f + g, fg are absolutely continuous. What about 11 f? Thus, if AC[a, b] denotes the space of all absolutely continuous functions, AC[a, b] is a vector subspace of BV[a, b]. Exercise 3. Show the composition of absolutely continuous functions needn't be absolutely continuous. Exercise 4. If f and 9 are absolutely continuous and 9 continuous.

i, show fog is absolutely

Exercise 5. If f satisfies a Lipschitz condition and 9 is absolutely continuous, show fog is absolutely continuous. Exercise 6. Is the uniform limit of absolutely continuous functions necessarily solutely continuous? Exercise 7. Show f(t) tl/3, O:S; satisfy a Lipschitz condition. Exercise 8. Let f : constant. Exercise 9. Let

f,

la, bJ

{b

Exereise 10. If f

< 1, is absolutely continuous but does uot

R be absolutely continuous. If f'

9 be absolutely continuous on

Ja

J: Ijldm.

--+

JiJdm

[a, b]

ab~

la, b].

0 a.e., show f

Show

(b .

+ Ja fgdm = f(b)g(b) - f(a)g(a).

--+

R is absolutely continuous, show Var(f

[a, b])

Exercise 11. Let f : [a, bJ --+ R. Show f is absolutely continuous if and only if x --+ Var(f : [a, xl) is absolutely continuous. Exercise 12. Let f : [a, b] --+ R. Show that f satisfies a Lipschitz condition as in Example 2 if and only if f is absolutely continuous and 1f'1 :s; L a.e. Exercise 13. Let f(x) l/vIx for x > 0 and frO) = 0, get) t2sin(Ilt) for t =f 0 and g(O) O. Show f is integrable over [g(O), g( 11' /2)], 9 is absolutely continuous but (1) fails for [a, bl = [0,11' /2J.

Chapter 5 Introduction to Functional Analysis 5.1

Normed Linear Spaces

In part 6 of these notes we consider some of the classic spaces of functions, most of which are associated with either measures or integrable functions. We begin by establishing an abstract framework, essentially due to S. Banach ([Bl]), in which we can study these function spaces. Let X be a vector space over the field F of either real or complex numbers. A topology T on X is a vector topology or linear topology if the maps (x, y) -+ (x + y) from X x X into X and (t, x) -+ tx from F x X into X are continuous [with respect to the product topologies]. If T is a vector topology on X, the pair (X, T) is called a topological vector space (TVS) or X is called a TVS if the topology is understood. A (semi)- metric linear space is a TVS whose topology is given by a (semi-) metric. For an example of a semi-metric linear space, we have Example 1 Let p.. be a finite measure

OIl

a u-aIgebra 1: of subsets of S.

Let LO(p..) be the space of all real-valued 1:-measurable functions; if m is Lebesgue measure on an interval I, we set LO(I)= LO(m). Then d(J,g)

f If gl d is 1 + If - gl p..

defines a semi-metric on LO(p..) [3.6.8]. Since convergence in d is exactly convergence in p..-measure [3.6.9], LO(p..) is a (complete) semi-metric linear space under d. Most of the spaces which we consider are normed spaces, we use the semi-metric linear space in Example 1 and the one in Example 8 to illustrate some of the important properties of normed spaces. Definition 2 A norm on X is a function (i)

IIxll

~ 0 for all x E X and

Ilxll =

1111 :

X -+ R satisfying

0 if and only if x

= 01 \

165

166

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Ilt:[11 = Itlllxli for all x X and t E F. (iii) Ilx + yl <::: Ilxll Ilyll for' all y E X [triangle If a funclion 1111 : X R 8(UI8Jtl~;" (ii), (iii) and (ii)

I!xll ?: 0 for all EX, then 1111 is called a semi-norm

(if)

on X [noif: from (ii) that

11011

= OJ.

We establish an important inequality associated with the triangle inequality. Let Ilxli yll + Ilyll by (iii) so Ilxli -Ilyll <::: II:/: yll. By symmetry :rll Ilx yll by (ii). Hence,

x, y E X. Theil lIyll-llxli <::: Ily

11;[ - yll ?: 111·1'1 Ilylll· if 1111 is a (semi-) llorlIl Oll

(iv)

X, the pair (X, III) is called a norm is understood, X is called a [serni- 'I LS or 'I LS]; if the If X is a (semi-) then. y) 113: . . . yll defines a metric 011 X which is translation invariant in the sellse that d(x + z,y + z) = y) for .1', y, E X. We always assnrrw that a NLS is equipped with the topology indnced the (semi-) metric d. We now show that a (semi)- '1LS is a TV::;.

Proposition 3 Let X be a scmi-NLS. Then tx fl'Orn F

(a) the map (t,.r) (b) the map (x, y)

(e)

the map x

-+

--t :1:

Ilxll

+ y from

fr'OTf! X

Proof: (aJ: Let Itk

and (a) follows. (b) follows from thf'

X

to

X

X x

X is continuous,

x+

X

is continuous,

R is continuous.

fl> 0 and

- [II

-+

O. Then

and (c) follows from (iv).

if it is complete uudf'f t.he semi-metric induced A semi-'1LS is said to he by the semi-norm, i.e., if every sequence converges. A completf' '1LS is called a Banach space or a B-8pace. vVe an interest.ing characteri7,ation for completeness in terms of series. Xk be a (formal) series in a semi-NLS X. The series Xk is said to Let

f

k=!

('onl'e7~ge

k

in X if the sequence of partial sums,

bk =

L J=1

xk

= lim Sk

in X if

f

k=J

xJ'

converges in

for t he sum of t he series. The series is said to be absolutely

Ilxkll < OC.

we write

5.1. NORMED LINEAR SPACES

167

Theorem 4 A semi-NLS X is complete if and only if eve7'y absolutely convergent series in X is convergent. Proof: =?: Let

I: IIxkll

<

00

and

t

= ;=1 Xi.

Sk

k=1

If k > j, then by the triangle

inequality

so {sd is a Cauchy sequence and must converge to some point in X. {=: Let {xd be a Cauchy sequence in X. Pick a subsequence {xn.} satisfying

Ilx

nk

+1

Xnk

I

< 1/2k. Since the series

f:

k=1

(X nk +1 -

X n .)

is absolutely convergent, it

converges to an element x E X so I

and the subsequence {Ink} converges to x + x n ,. Since {Ik} is Cauchy, Ik --t I + X n ! ' We now give several examples of metric linear and norrned spaces. Most of the spaces which we consider are normed spaces, but we give two examples (Examples 1 and 8) of metric linear spaces in order to illustrate some of the important properties of normed spaces. Example 5 If S =f. 0, let B(S) be the space of all bounded, real-valued functions defined on S. B( S) is a vector space when addition and scalar multiplication of functions is defined pointwise. B(S) also has a natural norm, called the sup-norm, defined by IIfll = sup{lf(t)1 : t E S} [Exercise 1]. B(S) is a B-space under this norm, for suppose that Ud is a Cauchy sequence in B(S). For t E S, Ifk(t) - fAt)1 :s: IIIk hll so Uk(t)} is a Cauchy sequence in R. Let f(t) limfk(t). We claim that f E B(8) and IIfk - fll --t O. Let E > O. There exists N such that k, j 2:: N implies Ilfk fjll < €. Then Ifk(t) fAt)1 < € for k, j 2:: Nand t E S. Letting j --t 00 gives l/k(t) f(t)l:s: E for k 2:: N, t E S so Ilfk fll:S: (for k 2:: N. Hence, fk - f E B(S) so f B(S) and Ik --t f with respect to the sup-norm. Example 6 Let S be a compact Hausdorff space and C(S) the space of all continuous functions on S. Then C(S) is a linear subspace of B(S). Since convergence in the sup-norm is exactly uniform convergence on S, C(S) is a closed subspace of B(S) and is, therefore, complete. Example 7 Let Ji be a measure on the O'-algebra, of subsets of S. As in §3.5 let L1(Ji) be the vector space of all real-valued, Ji-integrable functions. Then IIflll Is If I dJi defines a semi-norm OIl P(Ji) which is complete [Riesz-Fischer Theorem].

If 8 = Nand Ji is counting measure on N, we set LI (Ji) of all sequences {t j} such that

lI{tJJIII = L Itil < 00. j=1

£.1. Thus, pI consists

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

168

We will consider other spaces of integrable functions in §6.1. The examples which we give now are sequence spaces; we give further examples of function spaces and study them in detail in later chapters. Example 8 Let s be the space of all real-valued (or complex-valued) sequences. Then s is a vector space under coordinate-wise addition and scalar multiplication. We define a metric, called the Frechet metric, by

d({$j},{tJ)=LI.5j

tjl/(1+ ISj

tjl)2j.

j=!

It is easily checked that d is a translation invariant metric on the triangle inequality].

.5

[Lpmma 3.6.7 gives

We first observe that convergence in the metric d is just coordinatewise convergence. For suppose that x" = {xj}:1 and x {Xj} are sequences in s. If

d( xk, x)

->

0, then lim"

Conversely, suppose

f:

1/2i

< (/2.

xj

xj for each j since

li£llxj = Xj

for each j and let

f

> O. There exists N such that.

There exists M such that k ::::: M implies

j=N N-I

L

Ixj ~ x;1 < f/2.

j=1

Hence, if k ::::: M, then 00

N-l

d(x\x)::;

L Ix 7-

Xii +

j

1/2 <

to.

j=1

It follows from this observation that

S

is complete under d [Exercise 2].

Example 9 Let [00 be the linear subspace of s which consists of the bounded sequences [so [00 = B(N)]. Then

defines a norm on [00 called the sup-norm and fOO is a B-space under the sup-norm (Example 5). Example 10 c is the subspace of [00 which consists of the convergent sequences. We assume that c is equipped with the sup-norm. We can show that c is a Bspace by showing that it is a closed subspace of the complete space [eo. So suppose = is a sequence in c which converges to the point j} E [00. Let

xl<

{xj}:1

x {x

5.1. NORMED LINEAR SPACES

lirxj = £k

for each k. Since Ilxk - xll

169

~ Ixj - Xjl for all j, lifIxj = Xj uniformly

oo

for j EN. Hence,

· I·Imx k· I1m J k

j

n = I·1m I·Imx k. = I·Imx · = I·Im"k J k j k J j

so x E c. Example 11 Co is the subspace of c consisting of all the sequences which converge to o. We assume that Co is equipped with the sup-norm. As in Example 10,

Co

is a B-space [Exercise 3].

Example 12 Let Coo be the subspace of Co which consists of all sequences x = {Xj} which are eventually 0, i.e., there exists N (depending on x) such that x j = 0 for j > N. We assume that Coo is equipped with the sup-norm. We show that Coo is not complete. Let f.k be the sequence in Coo which has a 1 in the kth coordinate and 0 in the other coordinates. If k

xk = L(1/j)e j , j=1

then {xk} is a Cauchy sequence in

Coo

which does not converge to a point in

Coo.

Coo is a very useful space for giving counterexamples to results involving completeness. We will give further examples of sequence spaces in §6.1.

Example 13 Consider Rn (or en). Rn has, of course, the usual Euclidean norm n

II x l1 2 = II(xI, ... , xn )1I2 =

L lxi, i=l

but it also has other natural norms. For example, n

Ilxlll = II(xI, ... , xn)111 =

L

IXil

i=l

or

[We give a further family of norms on Rn in §6.1.] If 1111I and 11112 are two norms on a vector space X, 1111I and 11112 are equivalent if there exist a, b > 0 such that a 11111 ::::: 11112 ::::: b 11111. We show that all norms on Rn are equivalent so, in particular, the three norms given above in Example 13 are equivalent. Theorem 14 Any two norms on Rn are equivalent.

170

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Theorem

14 Any two norms on

Rn

are equivalent.

Proof: By Exercise 5, it suffices to show that any norm, 1111, on Rn is equivalent to the Euclidean norm, Illb. Let be the vector in Rn which has a 1 in the ith coordinate and 0 in the other coordinates. If x = (XI, ... ,x n ) ERn, then

c,

by the Cauchy-Schwarz inequality. Next, let S = {x : IIxll2 I} and define f : S R by f(x) = Ilxll. By the inequality above the identity (Rn,11112) -> (Rn, IIID is continuous and x -+ IIxll is continuous from (Rn, 1111) to R [Proposition 3] so f is continuous with respect to 11112' Since S is compact with respect to 11112' f attains its minimum on S, say at Xo. Note that f(xo} = Ilxoll = m > O. If x ERn, x =fi 0, then xl IIxll2 E S so

That is, Ilxll 2 m Ilx112. For an example of two norms which are not equivalent, consider thc sup-norm,

111100 on If xi

and the norm,

Coo

tt

IIxll!

=

.f Ixd, where x

{x;} [notc this is a finite sum].

1::=1

ei , then

IIxilloo

Iii

-+

0 while

Ilxilil

1 [Exercise 6].

)=1

An interesting consequence of Theorem 14 is

Corollary 15

Any finite dimensional subspace of a NLS is closed.

Proof:

By Theorem 14 any finite dimensional subspace is complete 6]. In R n it is known that sets are compact if and only if they are closed and bounded. We now show that this condition characterizes finite dimensional NLS. For this we require a lemma of F. Riesz.

Lemma 16 (Riesz) Let X be a NLS and Xo a proper, closed subspace of X. Then for every 0 < () < 1, there exists xe E X such that Ilxeli = 1 and IIx xell 2 for every x E Xo.

°

Proof: closed, d Xe

= inf {llx XIII: x E Xo}. Since Xo is E Xo such that IIxI xoll::S dlO since dlO > d. Set xo)1 IIxI xoll· Then IIxeli = 1 and if x E Xc, Ilxo - XIII X + Xo E Xo so Let

Xl

E

X\Xo and set d

> O. There exists

(XI -

Xo

11- .,.,--1~ IIClixl - xoll

IIx xoll = Ilx - -:-:---'--;-;- + IIXI Xo - xoll -

x

+ xo)

xIii

171

5.1. NORMED LINEAR SPACES

Example 17 In general, Xe cannot be chosen to be distance 1 from Xo although this is the case if Xo is finite dimensional; see Exercise 17. Consider C[O, 1] equipped with the sup-norm. Let X be the subspace of C[0,1] consisting of those functions x satisfying x(O) = 0 and Xo = {x EX: IJ x(t)dt = O}. Suppose there exists XI E X such that IIxIl1 1 and Ilx xIII ~ 1 for all X E Xo. For y E X\Xo, let e = IJ xI! y. Then XI - ey E Xo so 1 Ilxl - (Xl - cy)1I = leillyll which implies

:s

Id

Now we can make IIJ yl as close to 1 as we please and still have lIyll

k

-+ 00

will work]. Thus, 1

:s lId XII.

= 1 [Yk(t) =

lId XII < l. {x EX: Ilxll :s I}

But, since IlxIIi = 1 and Xl (0)

Theorem 18 Let X be a NLS and suppose that the unit ball B is compact. Then X is finite dimensional.

tllk as

0,

Proof: Suppose X is not finite dimensional. Let 0 # Xl E B and set Xl = span{xd. Then Xl C X and XI is closed by Corollary 15. By Lemma 17, there exists X2 E B such that IIx2 - XIII ~ 1/2. Let X 2 = span{xl,x2}. Then X 2 is a proper, closed subspace of X so by Lemma 17 there exists X3 E B such that IIX3 x211 ~ 1/2, IIX3 - XIII ~ 1/2. Inductively, there exists a sequence {xd C B satisfying IIxi - Xjll ~ 1/2 for i # j. Hence, B is not sequentially compact. A subset B of a semi-NLS is bounded if sup{lIxll : X E B} < 00; this is equivalent to B being bounded in the semi-metric induced by the norm. From Theorems 14 and 18, we obtain Corollary 19 Let X be a NLS. Then X is finite dimensional bounded subsets of X are compact.

if and

only if closed,

The conclusion of Corollary 19 does not hold in a metric linear space [Exercise 10j. Exercise 1. Show the sup-norm defined in Example 5 is actually a norm and convergence in the sup-norm is exactly uniform convergence on S. Exercise 2. Show s is complete under the Frechet metric. Exercise 3. Show Co is a B-space under the sup norm. Exercise 4. Give an explicit example of a series in Coo which is absolutely convergent but not convergent. Exercise 5. Let IIlli (i = 1,2,3) be norms on a vector space X. If 111/1 is equivalent to 11112 and 11112 is equivalent to 11113' show 111/1 is equivalent to 11113· \

172

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Exercise 6. If 11111 and 11112 are equivalent norms, show they have the same convergent (Cauchy) sequences [generate the same topologies].

Exercise 7. Show that a subset B of a semi-NLS is bounded if and only if {xd and tk --t 0 implies tkxk --t O. Exercise 8. Show {e k

:

c B

kEN} is a closed, bounded set in C= which is not compact.

Exercise 9. A subset B of a metric linear space (TVS) is bounded if {xd --t 0 implies tkxk --t 0 [see Exercise 7]. Show a compact subset is bounded.

c

B,

tk

Exercise 10. Show that a subset B of s is bounded if and only if B is coordinatewise bounded. Show that a subset of s is compact if and only if it is closed and bounded. Hint: Use a diagonalization procedure.

Exercise 11. What is the closure of

coo

in C=?

Exercise 12. Show Coo, Co, c, CI are all separable.

Exercise 13. Show C= is not separable. Hint: Consider all sequences of O's and 1'so

Exercise 14. If M is a linear subspace of a NLS X which has non-empty interior, show M = X.

Exercise 15. If X is an infinite dimensional Banach space, show X has uncountable algebraic dimension. [Hint: If {xd c X, consider Fk = span{ XI, •.. ,xd and use the Baire Category Theorem.] Give an example of a NLS with countably infinite dimension.

Exercise 16. Let {Xj} C X, a B-space. Show {Xj} is bounded if and only if L,tjXj converges for every {t j} E C1 .

Exercise 17. Show that if Xo is finite dimensional in Lemma 16, then () can be taken to be equal to 1.

173

5.1. NOHMED LINEAR SPACES

Exercise 18. Let cs be the sequence space consisting of all sequences x = {x il such that

Xj

converges. Show cs is a Banach space under the norm

Exercise 19. Let X, Y be NLS. Show

define equivalent norms on X x Y. Exercise 20. Let X be all lEe [0, 1J such that I' E C [0, 1]. Show X is not complete with respect to the sup-norm but is complete with respect to the norm, 11111 = sup {II (tll : 0 t I} + sup {II' (tll : 0 t I}.

:s :s

:s :s

174

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

5.2

Linear Mappings between Normed Linear Spaces

In this section we consider the continuity of linear mappings between normed spaces. Let X and Y be semi-NLS and T : X -+ Y linear.

Proposition 1 The following are equivalent: (i) T is uniformly continuous, (ii) T is continuous, (iii) T is continuous at 0, (iv) there exists lv! 2: 0 such that (*)

IITxl1 ~ M Ilxll for all x EX.

Proof: Clearly (il =;. (ii) =;. (iii). (iii) =;. (iv): If (*) fails, there exist Xk X sllch that IITxk11 > k211xkll for each k. If Vk = xk/k Ilxkli, then Yk -+ 0 while I TVkl1 > k so (iii) fails. (iv) =;. (i): If (> 0, put (j f/(M + 1). If Ilx - yll < (j, then

IITx Tyll

111'(x- y)11 ~ M Ilx -

yll < c

The space of all continuous linear operators from X into Y is denoted by L(X, Y); L(X, Y) is a vector space under the operations of pointwise addition and scalar Illul· tiplication. If X =: Y, we write L(X,X). We define a semi-norm, called the operator norm, on L(X, Y) by sup{IITxll: IIxll ~ I}. Note that if T satisfies (*), then IITII ~ 1'.{ and 111'/1 is the infimum of all such numbers M satisfying C'l [n particular, we have 1) IITllllxll for x E X

Proposition 2 (i) The operator norm is a semi-norm on L(X, Y). (ii)

IITII =

0 if and only if IITxl1

0 for all x EX.

(iii) If Y is a NLS, then L(X, Y) is a NLS undEr the operator norm. (iv) If Z is a semi-NLS, T E L(X, Y), S E L(Y, Z), then ST E L(X, Z) and

!ISTII

~ IISIII/TII·

Concerning completeness, we have

Theorem 3 If Y is a complete NLS, then L(X, Y) is complete.

5.2. LINEAR MAPPINGS BETWEEN NORMED LINEAR SPACES

175

Proof: Suppose {Tk} is Cauchy with respect to the operator norm. If x E then (5.2) so {lkx} is Cauchy in Y. Let Tx lim Tkx. Then T: X -> Y is iinea.r. We show T E L( X, Y) and Tk -> T in the operator norm. Let I: > O. There exists N such that k, j ::: N implies 1111, lill < (. Let j -> 00 in (2) so IITkx Txll"::: E !Ixll for k ::: N and x E X. Hence, II, T E L(X, Y) with IITk - TIl . : : E for k N so T E L(X, Y) and Tk ...... T in operator norm. We establish the converse of Theorem 3 in 5.6.1.7. If X is a metric linear space (TVS), a linear map from X into the scalar field is called a linear fu.nctional. The dual of X is the space of all continuous linear functionals on X and is denoted by ,i.e., X' = L(X,F). If x' is a linear functional on X, we often write x'(x) (x', x) for x E X. If X is a semi-NLS, it follows from Theorem 3 that X' is a B~space under the operator norm Ilx'll sup{l(x' , x)1 : Ilxll <;:; I}; this norm is called the dual n()rm or

X'. If X and Yare semi-NLS, a map U: X -> Y is an isometry if IlUx - Uyll = for x, y E X, i.e., if U preserves distances. If X and Yare NLS, t.hen X and Yare linw1'ly isometric if there is a linear isomet,ry from X onto Y; if X and Yare linearly isometric, it is customary to write X = Y. In later chapters we describe the duals of many classic function spaces. As an example, we now describe the dual of the sequence space Co. F. Riesz gave dcscript,ions of IIlany of the duals of classic function spaces so any result which describes the dual of a specific function space is often referred to as a "Riesz Represent.ation Theorem".

Ilx - yll

Example 4 c~ and [I are linearly isometric, c~

Let

f

E c~. Set Yk

(1, ek)

[I.

for kEN and set y = {yd. We claim that y E £1, If n

= :L(signYk)ek,

Xn

k=1

then

n

(1, Xn) = :L IYkl <;:;

IIfllllXnll oo

..:::

IIfll

k=1

00

(J, x)

:L Xk

(1, ek) =

k=1

and

00

1(J,x}I"::: IIxll oo which implies

Ilfll

<;:;

IlylI l ·

Hence,

IIfll

=

Ilyill'

XkYk

176

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

f

Thus, the map U which sends

-->

c~

Y is an isometry from

obviously linear. We show U is onto fl. If Y

= {Yd

E f1, (fy, x)

=

into f1 which is

I: XkYk defines a k=l

continuous linear functional fy on Co and U(fy) = Y so U is onto. We give a similar description of the dual of fl. Example 5 Let f E (fl)'. Set Yk

= (f, ek ) and Y = {yd. Then Y

IIYlloo 'S Ilfll since l(f,ek)1 'S Ilfll· If x = {xd E fl, then x =

k=1 and

If(x)1 'S

E foo with

I: Xkek so

k=1

k=1

II{Ydlloo Ilxll l · Hence Ilfll 'S II{ydIL", and Ilfll = II{Ydlloo'

Thus, the map U which associates f E (fl)' with Y E foo is an isometry from (fl)' into foo which is obviously linear. Now U is actually onto f= since if Y = {yd E f oo ,

(fy, x) =

I:

YkXk defines a continuous linear functional fy on fl with U(fy) = y. k=1 Hence, (fl)' = f=. A more general version of this result is given in Theorem 6.2.4. Exercise 1. If X, Yare semi-NLS and T : X --> Y is linear, show T is continuous if and only if T carries bounded subsets of X into bounded subsets of Y. Exercise 2. Define R, L : fl

-->

fl (or f=

-->

R{tj} = {O, t l , t 2, ... } L{t J } = {t2,t3, ... }

f=) by (right shift), (left shift).

Show Rand L are continuous and compute IIRII, IITII. Exercise 3. Let XI be a dense linear subspace of the NLS X and let Y be a Bspace. If T : Xl --> Y is linear, continuous, show that T has a unique linear extension f' E L(X, Y) with IITII = 11f'11. In particular, X; = X' [equality here means linearly isometric]. Exercise 4. Describe the dual of

Coo.

Hint: Exercise 3 and Example 4.

Exercise 5. Let X be a NLS. Show X and L(F, X) are linearly isometric. Exercise 6. Let X be a B-space and T E L(X). Define eT and show eT E L(X). If S, T E L(X) commute, show eT +S = eT eS .

5.2. LINEAR MAPPINGS BETWEEN NORMED LINEAR SPACES

177

Exercise 7. Show that Co and c are linearly homeomorphic. Exercise 8. Show that c' and fl are linearly isometric under the correspondence which associates with each y {y.} E fl the linear functional fy E c' defined by 00

(Ill) {x.}) =

I: YI:XI:_I 1:=1

where Xn

=

limxk.

Exercise 9. Show the linear functional L : c continuous and compute II LII.

~

R defined by L{ xd

Exercise 10. Show that Sf and Coo are algebraically isomorphic under the map which associates with each Y {y.} E Coo the linear functional fy on s defined by

Exercise 11. Define f : Coo

->

R by f({xd)

Xk.

Show f is linear but not

continuous.

Exercise 12. Let {t,} E s. For {s;} E s, set T{s;} = {tiS;}. Find necessary and sufficient conditions on {t;} so that TEL (£1,£1) or T E L((,oo,£oo). Compute IITII. Exercise 13. Show that any linear map from Rn into a semi-NLS is continuous. Exercise 14. Describe the dual of Rn (including the dual norm) for each of the norms

II 112' II III' II 1100'

Exercise 15. let k : [0,1) x[O, 1)

->

R be continuous. Show K f (s)

l

k(s, t)f (t) dt

defines a continuous linear operator from C[O, 1] into C [0,1]. Show K carries bounded subsets of C [0, 1) into relatively compact subsets. (Hint: Arzela-Ascoli.)

Exercise 16. Let a!; > 0, ak 1 0, a = {a!;}. Define a norm by Co by II{tdlia = sup {Itkakl : k}. Describe the dual of Co under II Iia and describe the dual norm of

II lIa'

178

5.3

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

The Uniform Boundedness Principle

In this section we establish one of the earliest abstract results in functional analysis, the Uniform Boundedness Principle (UBP), and one of its most important consequences, the Banach-Steinhaus Theorem. The Uniform Boundedness Principle is one of the three basic abstract results in functional analysis, along with the Closed Graph/Open Mapping Theorems and the Hahn-Banach Theorem; these theorelTls will be discussed in later sections. Let X, Y be semi-NLS. The proof of the UBP which we give is based on a technique called a "gliding hump" or "sliding hump" argument. Theorem 1 (UBP) Let X be complete. If F c L(X, Y) is pointwise bou.nded on X [i.e., {Tx: T E F} is bounded for each x E Xj, then {IITII : T E F} is bounded. Proof: If the conclusion fails, there is a sequence Pi} c F satisfyillg 11'1;11 > i'22i for each i. Then for each i there exists Xi E X, Ilxill :S 1, such that II'Fixill > i'22'. For convenience of notation, set Si = 2- i T i and Zi = 2-ixi. Then Ilzill :S 2-', 118i zi l > i, lim SiZj = 0 for each i and lim SiZJ = 0 for each j by the pointwise bOllll(kdlleSS J

'

assumption. Thus, the rows and columns of the matrix M = [S,Zj] converge to 0 so there is a subsequence {nil such that IISn,Zn,11 2- i- j for i f= j [see Lemma 2.8.1 and its proof]. Then for each i,

:s:

f J=l

iti

IISn,Zn, I :s

f

Ti-j = Ti

J=l

and IISn,zn.11 > ni, i.e., for each i the sequence {Sn.ZnJ~1 has a "hump" in the ith coordinate and the sum of the norms of the other elements in the sequence is much smaller than the "hump" and as i increases the "hump" slides to the right since {n J } is increasing. We now "gather" or "collect" the points {znJ which give rise to the (X)

humps by setting Z =

I: Zn ; j=l

note that this series converges in X since the series is

'

absolutely convergent and X is complete [5.l.4]. We now have

IISn.ZII =

II~ Sn.zn, 112: IISn.zn.11 - ~ I Sn.zn, 112: ni -

Ti.

iti

But, since {Tiz} is bounded, {SiZ} = {2-iTi z} should converge to 0 contradicting the inequality above. Without some condition, such as completeness, on the domain space X the UBP can fail.

5.3. THE UNIFORM BOUNDEDNESS PRINCIPLE

179

Example 2 Define" : Coo R by Ii( {t j}) = it i . Then fi is a continuous linear functional on Coo with = i, and the sequence {fi} is pointwise bounded on C-{lQ.

'lfd'

There are versions of the UBP which hold without any hypothesis on the domain space. For a discussion of such results see [AS], [Sw1], or [Sw2j. We derive an important consequence of the UBP concerning the continuity of the pointwise limit of a sequence of continuous linear operators.

Theorem 3 (Banach-Steinhaus) Let X be complete, Y a NLS and {Td C L(X, V). if lim Tkx Tx exists for each x EX! then T : X --t Y is linear and continno1LS. Proof: T is clearly linear. The sequence {lk} is pointwise bounded on X so by the UBP sup{ IITk I : k} M < 00. Thus,

IITxll so

TE

L(X, Y) with

lim II TkX II

:::: M IIxil

for x EX

IITII :::: M,

Again without some aS8umpt,ion on X, this result can fail.

Example 4 Define

It ; Coo

--t

R by

M {tj})

=

t j . Each" is continuous and for

{tj} E Coo, fi({lJ})

--t

I)i = f({lj}), j=1

but

f

is not continuous (Exer. 5,2.11).

We give an application of the Banach-Steinhaus Theorem to sequence spaces also Exercise 3]. A further application to Fourier series is given in Chapter 6.6.

Proposition 5 Suppose the sequence {tj} is such that

f

tjSi converges for every

j=l

Proof: For each k define Jk E c~ = i 1 [5.2.4 J by h( {Sj} )

for {sil E Co. By Theorem 3,

J E (Co)'.

By Example 5.2.4 {tj} E i

For a discussion of the evolution of the UBP see [Sw1].

l

(Co)'.

180

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Exercise 1. Let Fe L(X, Y). Show {1IT11 : T E F} is bounded if and only if F is uniformly bounded on bounded subsets of X if and only if F is equicontinuous at o if and only if F is equicontinuous on X. [F is equicontinuous at Xo E X if for every f > 0 there exists 8 > 0 such that Ilx - xoll < 8 implies IITx Txoll < f for all TEFl Exercise 2. In Theorem 3, show

IITII :SlimllTkll < 00.

Exercise 3. Suppose the sequence {tj} is such that

f

tjSj converges for every

j=1

Exercise 4. Let {T,d C L(X, Y) be equicontinuous and let Z be a dense linear subspace of X. Show that if Y is complete and limTkz exists for each z E Z, then lim11x Tx exists for each x E X and T E L(X, Y). Exercise 5. Let X, Y, Z be NLS with X a B-space. Let B : X x Y --> Z be a separately continuous bilinear map. Show B is jointly continuous [use Exer. 5.1.19]. Hint: Use the UBP to show there is a constant b such that IIB(x,Y)11 bllxilllYIi.

:s

Exercise 6. Give an example showing completeness cannot be dropped in Exercise 5. Exercise 7. Use the Baire Category Theorem to prove the UBP. Hint: Consider

Fk

{x: s~PIITnxll:S k}.

,5.4. QUOTIENT SPACES

5.4

181

Quotient Spaces

In this brief section we consider the quotient of a NLS. These results are used later in dealing with some of the classical spaces of functions. Let X be a semi-NLS and M a linear subspace of X. If x E X, we denote the coset x + Min X/M by [xJ x + M. We define a semi-norm on X/M by lI[xlll'

inf{lIx

+ mil: m

distance(x, M)).

E M}

(5.1 )

Proposition 1 (i) IIII' is a semi-norm on X/M. (ii) X/M is a NLS

{:?

M is closed.

(iii) The quotient map x

-+

[xl from X onto X/M is norm reducing [and, therefore,

continuous} and open.

(iv) X/M is complete if X is complete. (v) If X is complete and M is closed, X/M is a B-space. Proof: (i): II[txlll' = inf{lltx + mil: m E M} It I 11 [xl II' for t :f O. II[xl

+ [ylll'

:s

inf{ltlllx

+ m/tll

:m E

llyI}

=

inf{lIx + y + m) + m211 : mi E M} inf{llx + mIll: ml EM} + inf{lIy + m211 : m2 E AI} !I [xl II' + lI[y]II'·

(ii): IHx]1I' = dist(x, M) = 0 if and only if x E M and [xl 0 if and only if x E M. (iii): Clearly IIxll ? Illxlll'. To show the quotient map is open we show that {x: IIxll < I} is mapped onto {[x] : II[xlll' < I}. Let II[xlll' 1 (j < 1. There exists m E M such that Ilx + mil < 1 and [x + m] [xl. (iv): Let

Z[Xk] be an absolutely convergent series in X/M.

For each k choose

k=1

mk E Al such that Then

f= Ilxk + mkll :s f= (11[xk]II' + 1/21:) <

1:=)

so

00

k=!

Z(Xk + mk) is absolutely convergent in

X and, therefore, convergent to some

k=1

x E X [Theorem 5.1.1]. By (iii), 00

[xl = I:[Xk 1:=1

so X/AI is complete by Theorem 5.1.4. (v) follows from (ii) and (iv).

00

+ mkl == I:[Xk] k=!

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

182

Proposition 2 Let K(X) sl1bspace of X.

{x EX:

Ilxll

= o}. Then K(X) is a closed, linwr

Proof: If x, y EK(X), then Ilix + sY11 ~ Illllxli + Isillyll = 0 so J«X) is a linear subspace. K(X) is closed by Proposition 5.1.3. Proposition 3 The ql10tient map X etry}. Proof: For

mE K(X), Ilxll

II·TII

->

XI I«X) is norm preserving {i.e., an isom-

11

mI

~

Ilx - mil

~

Ilxll

80

Ilxll

IHxlll'·

Corollary 4 XI K(X) is a NLS and if X is complete, XI K(X) is a B-space. As an example, consider V (iJ) with the V-norm, I Ill' If [( I< (V (fl)), then E K if and only if f = 0 Jl-a.e. Thus, the coaets of V (Jl) I I< comist of equivalence classes of functions which are equal Jl-a.c.

f

Exercise 1. Let X, Y be scmi-NLS and l' E L(X, Y). Show the ind1lced map T: XI kerr -> Y is continuous and 111'11 =

Ilil

5.5.

THE CLOSED GRAPH/OPEN MAPPING THEOREMS

5.5

183

The Closed Graph/Open Mapping Theorems

In this chapter we discuss another of the three basic principles of functional analysis, the Closed Graph Theorem (CGT) and its companion the Open Mapping Theorem (OMT). Let X, Y be NLS and T: X -+ Y linear. T is said to be closed if its graph {(x, Tx) : x E X} is closed in X X Y. Thus, T is closed if and only if Xk -+ x in X and TXk y in Y implies that y Tx. Any continuous linear operator T E L(X, Y) is obviously closed; however, as the following example shows not every closed operator is continuous.

Example 1 Let X = {f E C[O, I] ; f' exists and is continuous on [0, I:} and assume that X is equipped with the sup-norm. Let Y C[O, 1) and D : X -+ Y be defined f'. Then D is obviously linear, closed [[DeS)lL7], but is not continuous by Df

[lltkll oo

1 while

II(tk)'lIex> = kJ.

Note that the domain space in this example is not complete. Indeed, the CGT asserts that if X and Yare complete, then any closed linear operator T : X -+ Y is continuous.

Theorem 2 (CGT) Let X I Y be Ba.nach spa.ces and T : X closed, then T is continuous. Proof: Set Z

{x EX:

IITxl1 <

-+

Y linear. 1f T is

I}. Since T is linear, X

kZ so from

the Baire Category Theorem (A2) some =: kZ contains an interior point. Hence, {y: Ily - xii < r} = S(x,r) C Z for some x E Z, r > O. If Ilzll < r, then

z so S(O, 1') C

Z.

1

1

2(x+z)

2(x-z)E

1-

1-

2

2

Z-

Z

-.

Z

Hence,

S(O, ar) C aZ

for every a

> O.

(5.1)

We now claim that if Ilzll < r, then IITzll 2. Since z E Z, by (1) there exists Xl E Z such that liz - XIII < r/2. Since Xl - z E 0) by (1), there exists X2 E such that liz Xl - x211 < 1'/2 2 • Continuing inductively produces a sequence

{xd C Z with

liz -

so that z = limsk = k

XI -

f

j=1

X2

Xi and

••• -

xkll < r/2k and Xk E (1/2 k -

IITxkl1 :; 1/2k - 1 •

L IITxill:; '=jH

)

z.

Put Sk =

Now {Tsd is Cauchy in Y since

k

IITsk - Tsjll:;

1

1/2.- 1 <

Xj

184 for k and

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALy"SIS

> j. Therefore, there exists y

IIT z l1

lIyll =

liE

E Y such that T Sk -> y. Since l' is closed, T z

TXkl1 ::;

E

IITxkll ::;

E

1/2

l

k -

y

= 2.

If IIxll ::; 1, then by the above IIT(rx/2)1I ::; 2 or IITxlI .1/r. lIence,"J' is continuous with IITII ::; 4/r. H one wants to prove that a linear map l' : X -> Y is continuous, one or t.lle usual procedures is to take a sequence {xd in X which converges to some x X and show that the sequence {Txd converges to Tx. The advantage of the eel' is that we may now assume that the sequence {Txd is convergent, and we are then required to show that it converges to the proper value, namely, Tx. An example of such an application of the CeT is given in Example 6.2.9 also Exercise 3, Exercise 6.1.18 and the end of §6.2J. We use the CGT to derive its companion rcsult, the OpCII :v1a.pping Theorem (OMT). Theorem 3 Let X, Y be Banach spaces and T E L(X, yO). 1fT is onto, T is open. Proof: Let T be the induced map,r : X/kef T -> Y [Exercise 5.1.]]. Then 1-1, continuous and onto Y so r- I is closed [E,xercise I]. By the CeT, is continuous so T is a homeomorphism. Since the quotient map X -> XI ker T is always open [5.4.1], T is open.

T is

Corollary 4 If X and Yare Banach spaces and T E L(X, Y)is 1-/ and onto, then l' is a homeomorphism. Corollary 5 Let X be a vector space with two complete norms 11111' 11112 on X. If 11111 is stronger than 11112 [i.e., induces a stronger topology}, then 11111 and 11112 are equivalent. Proof: The identity map (X, 11111) the resulL

->

(X, 11112) is continuous so Corollary 1 gives

The completeness of both norms in Corollary 5 is importa.nt, sec Exercise 2. Remark 6 The CGT is often derived from t.he OMT; sec [TL] for such a. development.

Exercise L Let T E L(X, Y) be 1-1, onto. Show 1'-1 is closed. Exercise 2. Show Corolla.ry 5 is false if both norms are not complete. [Hint: Consider G[O,l] with 111100 and 11/11 III (t)dt.]

f:

5.5. THE CLOSED GRAPH/OPEN MAPPING THEOREMS

185

Exercise 3. Let X, Y be Banach spaces and T : X Y linear. Suppose AcY' separates the points of Y. If y'T is continuous'll y' E A, show T is continuous. Exercise 4. Suppose that 1111 is a complete norm on e[0,1] such that implies Ik(t) I(l) for every l E [0,1]. Show 1111 is equivalent to 111100'

IIIk III

-t

0

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

186

5.6

The Hahn-Banach Theorem

In §5.6.1 we will study some of the relationships between a NLS and its dual space. In order to facilitate this study, we need a very important preliminary result called the Hahn-Banach Theorem. This is one of the three basic principles of functional analysis and has applications to a wide variety of problems. Besides using this result in 5.6.1 we give additional applications in sections 5.6.2 and 5.6.3. For an interesting discussion of the history of the Hahn-Banach Theorem see [Ho].

Definition 1 Let X be a vector space. A function p : X tional if

(i) p(x

+ Y)

(ii) p(tx)

R is a sublinear func-

::; p(x) + p(y) Vx, Y E X, tp(x) Vi

2: 0, x E X.

A semi-norm is obviously sublincar but not conversely. The Hahn-Banach Theorem guarantees that any linear functional defined Oil a subspace of a vector space which is dominated by a sublinear functional call be extended to a linear functional defined on t.he entire vector space and the extension is still dominated by the sublinear functional.

Theorem 2 (Hahn-Banach; real case) Lei X be a real vector space and p : X R a sub linear functional. Lei M be a linear subspace of X. If f : 1\1 - t R is a linear functional such that f(x) ::; p(x) Vx E M, then :3 a linear functional F : X - t R such that F(x) = f(x), Vx E M and F(x) p(x) Vx E X. Proof: Let E be the class of all linear extensions 9 of f such that. g(T) ::; p( x) Vx E V(g), the domain of g, with V(g) =2 At. Notc E fc 0 since fEE. PdfLial order E by 9 < h if and only if h is a linear extension of g. If C is a chain in E, then U gEE gEe

is clearly an upper bound for C so by Zorn's Lemma E has a maximal clement. F. The result follows if we can show V( F) X. Suppose :3XI E X\V(F). Let Atl be the linear subspace spanned by V(F) and Xl. Thus, if Y E Atb Y has a unique representation in the form y rn + tXJ where rn E V(F), t E R. If z R, then FI(y) = FI(rn

+ tTl)

F(m)

+ tz

defines a linear functional on MI which extends F. If we can show that it is possible to choose z such that PI(Y) ~ p(y) Vy E ]V!l, this will show PI E E and contradict the maximality of F.

5.6.

187

THE HAHN-BANACH THEOREM In order to have

we must have for t > 0,

or since mit E V(F), if z satisfies

z:::; -F(m) + p(m + Xl) Vm E V(F),

(5.2)

then (1) holds for t :::: O. For t < 0,

1

1

z :::: -TF(m) + TP(m

+ txd

F( -mit)

p( -mit - Xl),

or since -mit E V(F), if z satisfies,

z :::: F(m)

p(m

Xl) Vm E V(F),

(5.3)

then (1) holds. Thus, z must satisfy

i.e., we must have

But,

so (4) does hold. To obtain a complex form of the Hahn- Banach Theorem, we need the following interesting observation which shows how to write a complex linear functional in terms of its real part.

Lemma 3 (Bohnenblust-Sobczyk) Let X be a vector space over C. Suppose F

=

f + ig is a linear functional on X. Then for x E F(x) f(x) if(ix) and f : f(x) if(ix) X - t R is R-linear. Conversely, if f: X - t R is R-linear, then F(x) defines a C-linear functional on X.

Proof; f and 9 are clearly R-linear. Now F(ix) iF(x) implies f(ix) +ig(ix) = if(x) - g(x) so f(ix) = -g(x) and F(x) f(x) if(ix). The converse is easily checked.

188

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Theorem 4 (Hahn-Banach; complex case) Let X be a vector space and p: X -> R a semi-norm. Let M be a linear subspace 01 X and I : At -> F a linear lunctional. II I/(x)l ~ p(x) Vx E M, then I has a linear extension F : X - t F such that IF(x)1 ~ p(x) Vx E X. Proof: Suppose F = R. Then I(x) ~ I/(x)1 ~ p(x) Vx E l'vf so Theorem 2 implies a linear extension F : X - t R such that F(x) p(x) Vx E X But then F(-x) -F(x) ~ pC-x) = p(x) so IF(x)1 ~ p(x). Suppose F = C. Then RI, the real part of I, is an R-linear functional on X such that IRI(x)1 ~ lI(xll ~ p(x) Vx E M. By the first part, :I a real linear functional II : X -> R which extends RI and satisfies III (x)1 ~ p(:1') Vx E X. Set. F(x) hex) ifl(ix). Then F is C-linear and extends I by Lelllma 3. For x write F(x) = IF(x)1 cia. Then

Despite its very esoteric appearance we will see in the next three sections t.hat the Hahn-Banach Theorem has a surprisingly wide variety of applications.

Exercise 1. Give an example of a sublinear functional which is not a semi-norm. Exercise 2. Show p( {tj}) = lim(t l n

+ ... +tn)/n defines a 8ublinear fUllctional

on

5.6.

THE HAHN-BANACH THEOREM

5.6.1

189

Applications of the Hahn-Banach Theorem in NLS

We use the Hahn-Banach Theorem to derive important properties of the dual space of a NLS. We begin by establishing an important result on extending continuous linear functionals.

Theorem 1 Let X be a semi-NLS and M a linear subspace. 3x' E X' such that x' extends m' and IIx ' ll IIm'll.

11 m '

E M ' , then

Proof: Define a semi-norm p on X by p(x) Ilm'llllxli. Then l(m',x)1 -:; p(x) for x E M. By Theorem 5.6.4 there exists a linear function x' on X extending m' such that l(x',x)1 :::; p(x) Ilm'llllxll for all x E X. Hence, x' E X' and Ilx'll :::; ilm'll. Clearly Ilm'lI :::; Ilx'll· Theorem 1 can be used to establish several important resulLs on the existence of cont.inuous linear functionals on a semi-NLS. Theorem 2 Let M be a linear subspace 01 a NLS X. Suppose Xo E X is s'uch that distance (xo, M) d > O. Then there exists x~ E XI such that II x~ II 1J x~( M) 0 and (x~, xo) = d > O. Proof: Set Mo = span{M,xo}. Define a linear functional Ion Mo by I(m + txo) = td, where mE M, t E F. Then I(M) 0 and I(xo) = d. Also, I E M~ wit.h 11/11 S; 1 since if t of 0, mE M, then 11m Actually, IIIII

It I limit + xoll

+ txoll

:::::

It I d

I/(m

1 since there exists {md C lv1 with Ilmk d

I/(mk

and 11/11 ::::: 1. Now extend 11/11 = 1 by Theorem 1.

I

+ txo)l· xoll1 d so

xo)1 :::;lI/lIlImk - xoll111/11 d

to a continuous linear functional x' E X' with Ilx'll

Remark 3 Note that Theorem 2 is applicable if M is closed and

Xo

1:: 1\-1.

Corollary 4 Let X be a NLS and 0 of Xo EX. Then there exists xri E X' sllch that Ilx~11 = 1 and (x~, xo) Ilxoli. In particular, il x of y, there exists x~ E X' sllch that (x~,x) of (x~,y), i.e., the dual 01 X, X', separates the points 01 X. Proof: Set lv1

{OJ in Theorem 2.

Corollary 4 insures that the dual of a NLS is rich in continuous linear functionals. Such phenomena does not hold in general for metric linear spaces.

190

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Example I) Consider X = LO[O, IJ (Example 5.1.1). Suppose f is a non-zero linear functional on X. Let 'f' E X be such that (1,'f') =I O. Let II = [0,1/2], 12 = (1/2, IJ and set 9; CJ,'f'. Then 'f' 91 + 92 so either (1,9I) =I 0 or (1,92) =I O. Choose one and label it 'f'I. Note m{t : 'f'1(t) =I O} s:: 1/2. Continue this bisection procedure to produce a sequence {'f'j} C X such that (1, 'f'j) 1= 0 and m{t : 'f'j(t) =I O} s:: 1/2j • Put h j 'f'j/ (1, 'f'j) so (j, hj ) 1 for all j but h j --+ 0 in m-measure since m{t: hj(t) 1= O} s:: 1/2]. Thus, f is not continuous at O. Hence, the dual space of X is {OJ. The dual norm of an element x' in the dual of a NLS X is defined to be the supremum of the values I(x ' , x)1 as x varies overthe unit ball {x : IIxll s:: 1} of X. We can use Corollary 4 to establish a result dual to this; the norm of an clement. x E X can be found by computing the supremum of I(x', x} I as x' varies over the unit ball of X'. Corollary 6 For x E X, a NLS,

IIxll

sup{l(x/,x)l:

Proof: If x' E X', IIx'll s:: 1, then I(x', x)1 S 4 there exists x' E X' with IIx'll 1 and (x',x)

IIx'll s::

I}.

Ilxll. On the other hand, Ilxli.

by Corolliuy

As another application of Corollary 4 we establish the converse of Theorem 5.2.3.

Theorem 7 Let X, Y be NLS with X complete.

=I

{OJ. If L(X, Y) is complete, then Y is

Proof: Let {Yd be Cauchy in Y. Choose Xo E X, Ilxoll = 1. By Corollary ;[ there exists x~ E X' such that Ilx~11 = 1 and (x~, xo) = 1. Define Tk E L(X, Y) by Tkx = (x~, x) Yk. Then so

IITk -

Till

and {Td is Cauchy in L(X, Y). Suppose

s::

IIYk -

Yill

n --+ T in L(X, V).

Then

The Canonical Imbedding and Reflexivity: Let X be a NLS. Let X" be the dual of X' (with the dual norm) and assume that X" carries its dual norm from X'. X" is called the second dual or bidual of X. Each x E X induces an element x E X" defined by (X,X') (X',X) for x' E X'. x is obviously linear and by Corollary 6,

Ilxll =

sup{l(x' , x} I : Ilx'li S I} =

IIxll

5.6. THE HAHN-BANACH THEOREM

191

so £ E Xu and 11£11 = Ilxll. Thus, the map J x ; X -> Xu defined by Jxx = x is a linear isometry which imbeds X into its bidual Xu; J x is called the canonical map or canonical imbedding of X into its bidual, and if X is understood, we write J Jx . A NLS is called reflexive if JxX = X". Note from Theorem 5.2.3 any reflexive space must be a B-space. It should also be noted that for a B-space X to be reflexive, X and Xu must be linearly isometric under the canonical imbedding JXi R.C. James has given an example of a non-reflexive B-space X which is linearly isometric to its bidual X". Rn is obviously reflexive; examples of reflexive B-spaces are given in subsequent chapters. An example of a non-reflexive B-space is given below (Example 9). An interesting consequence of Corollary 6 is given by Corollary 8 Let X be a reflexive B-space. Then every continuous linear functional x' E X' attains its maximum on the unit ball of x. Proof: By Corollary 6 there exists x" E X" such that Ilx"ll 1 and (x", x') But there exists x E X such that Jxx x" so Ilxll 1 and (x', x) = Ilx'll.

Ilx'll.

It is an interesting result of James that the converse of Corollary 8 holds We can use Corollary 8 to give an example of a non-reflexive space.

Example 9

f

E

do

and

Corollary 8

Co

is not reflexive. Define f :

llill = Co

f

Co ->

R by f( {tj})

Ifj" However, there is no {tj} E

Co

with

j=1

[J]].

tj/j!. Then

tjli!

=

Ilfll.

By

is not reflexive.

As an application of the canonical imbedding, we have Theorem 10 Every NLS X is a dense subspace of a B-space a completion). Proof: Set isometry J x ].

X = JX

X

(i.e., every NLS has

C X" [where we are identifying X and J x X under the linear

As a further application of the canonical imbedding, we use the Uniform Boundedness Principle to derive a boundedness condition for NLS. Theorem 11 A subset B of a NLS X is bounded if and only if x'(B) is bounded for each x' E X'. Proof: =>: This follows from the inequality Ix'(x}1 :::: Ilx/llllxll. Let J be the canonical imbedding of X into its bidual. Then {Jb : b E B} is pointwise bounded on X', and since X' is complete, {IIJBII = Ilbll : b E B} is bounded by the UBP (5.3.1). {=;

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANAl_YSIS

192

Several additional properties of reflexive spaces are given in the exercises. Separability: We consider some separability results concerning a NLS and its dual. Theorem 12 If the dual of a NLS X is separable, then X is separable. Proof: Let {xU be dense in X'. For each k choose Xk E X such that Ilxkll 1 and l(xLxk)1 :::: IIx~II/2. The subspace Xl spanned by {xd is separable 8), and we claim that Xl is dense in X. If this is not the case, by Theorem 2, there exists x' E X', x' =I 0, such that (x', XI) O. There exists a subsequence {x~J converging to x'. Then

Letting k

-+

00

gives 0

Ilx/II/2 so x' =

OJ the desired contradiction.

The converse of Theorem 12 is false; see Exercise 6. However, for reflexive spaces, we have Corollary 13 Let X be a reflexive B-space. Then X is separable

~

X' is sepamble.

Proof: <¢=: Theorem 12. =*: J x X = X" is separable so X' is separable by Theorem 12. Exercise 1. Show that a B-space X is reflexive if and only if X' is reflexive. Exercise 2. Show that a closed linear subspace of a reflexive space is reflexive. Exercise 3. If X is reflexive and X' contains a countable set which separates the points of X, show X' is separable. Exercise 4. If X is a NLS, show JxX separates the points of X' Exercise 5. If X is reflexive, show X' has no proper closed subspaces which separate the points of X. Exercise 6. Show the converse of Theorem 12 is false. [Hint: Exercise 5.1.13.] Exercise 7. Let X, Y be NLS and T E L(X, V). Show that T'y' y'T defines a IIT'II. T' is called linear operator T' from Y' into X' which is continuous and Wfli the adjoint or transpose of T. Exercise 8. Let D be a countable subset of a NLS X. If XI is the snbspace spanned by D, show Xl and are separable.

5.6.

THE HAHN-BANACH THEOREM

5.6.2

193

Extension of Bounded, Finitely Additive Set Functions

We give an application, due to B.J. Pettis, of the Hahn-Banach Theorem to the extension of bounded, finitely additive set functions defined on algebras. Let A be an algebra of subsets of 5 and E the O"-algebra generated by A. Let 5(A) [5(E)] be the vector space of all real-valued A-simple [E-simple] functions; we assume that 5(A) [5(D] is equipped with the sup-norm, II'PII sup{I'P(t)l: t E Let Il : A -+ R be a bounded, finitely additive set function. We consider the possibility of extending Il to E. Let I : 5(A) -+ R be the linear functional induced by integration with respect to Il, (J,'P) = fs'Pdll [Remark 3.2.2]. Since 1(J,
J1 is bounded and gives a bounded, finitely additive extension of Il to the O"-algebra E generated by AIn contrast to the situation in §2.4 where we considered the extension of premeasures, a bounded, finitely additive set function defined on an algebra can have an infinite number of bounded, finitely additive extensions to the generating O"-algebra [see [HY]].

Exercise l. Show 11/11 = IIlI (5). Exercise 2. Show the existence of a bounded, finitely additive set function defined on a O"-algebra.

194

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANAL'I{SIS

5.6.3

A 'franslation Invariant, Finitely Additive Set Function

V\I'e show that the Hahn-Banach Tlwor<'Jl1 can be llsed to ;;JIOW that thc problem of measure discussed ill §1.3 has a, solut.ion. That is, we show Ih.> cxistpllcc of a nOllt.ranslation invariant, finit.ely additiv(' set fUlIct.ioll defined on the hounded subsets of R. W.. begin showing llw existcTlC<' of a Banach int('gral awl t.hen use Utc integra'! to construct. th" desir ..d translat.ion invilriallt, finitely a,dditiv(' set fund ion OIl the power set of R. Let P be the space of all bounded real-valued funct.ions ddined Oil R which have lwriod 1 equipped with the sup Ilorm, 11,1. For E P and I I, ... ,I" E R set

.r

u(.f:lh ... ,ln)="sup{

((I+lk)/n:IER) k,cl

and

p(.f) = i Jl f{ U(.f : ll' ' .. , I,,) : II,' , .• In E R}. Note p is finite since p(.f) <::: Ilfll"" oc. \Ve show that p is sublinear. elf'ady, 1) is pmit iv.' homogeneous. We show t.Ilat p is snhaddilivf'. Let f 0 and II, I2 E 1'. Pick "1.···,-'''' and 11,.' ,1" ouch that V.(fj:,sI • . . • Sm)

Put.

l'i}

=

Si

+ IJ

P(.fl)+(,

for i = 1, ... ,III, J

< .!.r:,sup n j=l

n(h:/I,.,.,t n )<

)j-(.

1.... . n. Theil

fdt+lJ+8,)/m:t}

l) so p(.h

+ h)

P(11)

+ p(Iz),

By the Hahn-Banach Theorem there exists 11 linear functional F: P - t R such that F(f) p(.f) for all f E P [define FOil {O} by F(O) = 0 and apply 5.6.2]. Note F( - f) -FU) .f) so p( -.f) <::: F(f)

p(.f) for f E P.

5.6. THE HAHN-BANACH THEOREM

195

Now F is positive [Le., F(f) ?: 0 for f ?: 0] since f ?: 0 implies p(f) ?: 0 and f) ~ 0 so by (2) F(f) ?: -p( - f) ?: O. Next, we claim that F is translation invariant. Fix hER. If f E P, set g(t) f(t + h) f(t) for t E R. We need to show that F(g) = O. Take tk = (k - 1)h for k = 1, ... , n + 1. Then

p( -

1

- - sup{J(t n+l

+ (n + 1)h) -

f(t) : t}

-+

0

as n -+ 00 so p(g) ~ O. Similarly, p( -g) ~ O. From (2), F(g) = 0 as desired. 1 and p( -1) -1. And, F is bounded since Also, F(1) = 1 from (2) since p(l) from (2), F(f) ~ p(f) ~ IIJII and

F(-f)

= -F(f) ~ p(-f)

~

II-fll = IIfll

~ llill and IIFI! ~ l. We now use F, the Banach integral, to induce a finitely additive set function. If E C [0, I), let kE be the periodic extension of CE to R and set ,,(E) = F(kE)' Since F is linear,,, is finitely additive on the power set, prO, 1), " is bounded (by 1) since IIFI! ~ 1, " is positive since F is positive and F(1) = ,,[0,1) = 1. Moreover, if A and B are subsets of [0, 1) such that A = B + t for some t, then ,,(A) ,,(B) hy the translation invariance of F, i.e., " is translation invariant in this sense. We extend" to the bounded subsets of R. First, consider any subset A of an interval of the form [j,j + 1). Then A - j C [0,1) so we may define ,,(A) to he ,,(A - j). If Be R is bounded, then B S [-n, n) for some n, and we may write

so

IF(f)1

n-l

B=

U

Bn[j,j+1)

i=-1l

and define

n-J

,,(B)

L

,,(Bn[j,j+1».

j=-n

It is easy to check that this extension of " has values in [0,00), is finitely additive, translation invariant, and ,,[0,1) = 1. Note that the commutativity of the group of translations was used ill the computation (1). This is important; Hausdorff has shown that there is no nOll-trivial, non-negative, fillitely additive set defined on the subsets of the unit sphere in R3 which is invariant under the (non-abelian) group of rotations on the sphere.

Exercise 1. Show the set function Ji( E) (,,( E) +,,( - E) )/2 has all of the properties of" and is also invariant under reflection. [Hence, Ji is invariant under the isometries of R ([Na] ).J Exercise 2. A Banach limit is a continuous linear functional L on Coo satisfying

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

196

(i) L( x) 2': 0 if x 2': 0 [i.e., if Xk 2': 0 for all k when x (ii) ifrx

=

(X2,X3,.")'

(iii) L«l, 1, ... ))

then L(rx) = L(x),

l.

S L(x) S Iimxk for x

(a) Show that if L is a Banach limit, e; > 0 choose n such that inf Xk so

Xk

+ e; -

{xo}],

Xn

::; Xn

< inf Xk

E fcc. Hint: For

+ e;

2': O. Use (i) and (iii) to show L (x) 2': inf Xk. Then use (ii).

(b) Show Banach limits exist. Hint: Use the functional p(x) = of Exercise 5.6.2 and the functional Lx = lim Xk for x E c.

Exercise 3. For A c N let IAI be the number of points in A. If L is a Banach limit, show p(A) = L(IAn (l, ... ,n}l/n) defines a bounded finitely additive set function on P(N). Exercise 4. Let A be an algebra of subsets of Sand T : S S such that 1'-1 A E A for A E A. Let v be a bounded, finitely additive set function on A. Let L be a Banach limit and define p on A by p (A) L (v (T-n A)). Show p is bounded, finitely additive and 1'-invariant (i.e., p(A) p(T-1A) for A E A).

5.7. ORDERED LINEARSPA.CES

5.7

197

Ordered Linear Spaces

As well as having natural norms or metrics, many of the classical spaces of functions also have natural orders. In this section we consider basic properties of ordered spaces. A partial order on a set E is a relation, satisfying the following properties:

(R) x :5 x holds for all x E E (reflexive), (A) if x :5 y and y :5 x, then x (T) if x

:5

y and y

y (anti-symmetry),

:5 z, then x :5 z (transitive).

We write y ~ x if and only if x :5 y. An ordered vector space is a real vector space X with a partial order, compat,ible with the algebraic operations, i.e.,

(i) if x :5 y and z E X, then x (ii) if x

:5 y and t

~

0, then tx

which is

+ z :5 y + z, :5 ty.

An element x of X is said to be positive if x X is denoted by X+.

~

0; the set of all positive dements of

Example 1 The space s has the natural order {xd :5 {Yk} if and only if Xk ~ Yk for all kEN. The subspaces ("-{lO, Co, C, £1 and £= inherit this natural order from s. More generally, we have Example 2 Let S :I 0 and F( S) the vector space of all real-valued functions defined on S [the operations of addition and scalar multiplication are defined pointwise]. If J, 9 E F(S), we define J :5 9 if and only if f(1) :5 g(t) for all t E S. Any vector subspace of F(S) is an ordered vector space under the order inherited from F(8). If X is an ordered vector space and x, y E X, then x and y have a supremum (infimum), denoted by x V y (x A y), if x :5 x V y, Y ~ x V y (x A Y ~ x, x A Y ~ y) and x ~ w, y ~ w (w ~ x, w ~ y) implies x V y :5 w (w ~ x A y). A vector lattice or a Riesz space is an ordered vector space X such that every two elements of X have a supremum and an infimum. We have the following elementary identities in a Riesz space.

Proposition 3 (i) x Vy = -[( -x) A (-y)], x A Y = -[( -x) V (-y)], (ii) xVy+z=(x+z)V(y+z),xAy+z

(x+z)A(y+z),

(iii) t(x Vy) = (tx) V (ty), t(x A y) = (tx) A (ty) for t

~

O.

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

198

Proof: We prove (ii) and leave the remaining statements for Exercise 2. Set a = x Vy + z, b (x + z) V (y + z). We show a b, b a. First, a - z = x V y implies x a z and y :::: a z so x + z a and y + z a. Hence, (x + z) V (y + z) b a. Next, b (x + z) V (y + z) so b:::- x + z and b:::- y + z or x b - z and y b z. Hence, x V y :::: b z or a = x V y + z b.

s:

s:

s:

s:

s:

s:

s: s:

s:

Remark 4 From (i) it follows that if X is an ordered vector space such t,hat every two elements of X has a supremum (infimum), then X is a Riesz space. If X is a Riesz space and x set x+ x V 0, the positive part of x, and set x- = (-x) V 0 (-x)+, the negative part of x. Then Ixl = x V (-x) is called the absolute value of x. We have the following properties:

Proposition 5 (i) x

(ii) Ixl = x+

+ x-

(iii) x+ 1\ x-

0,

x+

x- J

J

(iv) xVy=(x-y)++y,

(v) Ix

+ yl

:::: Ixl

+ Iyl,

s: Ix y+1 s: Ix

(vi) Ilxl-lyll (vii) Ix+ -

(viii) Ix V z - y V zl

yl, yl, (x

s:

Ix

+ y)+

:::: x+

yl, Ix 1\ z

(ix) Ixl :::: y if and only if -y

x

+ y+,

Y 1\ zl :::: Ix - yl,

y.

Proof: (i): By Proposition 3,

x-

+x

(-x) V 0 + x

0 V x = x+.

(ii): By Proposition 3 and (i),

Ixl

xV(-x)

(2x)VO

x

2x+-x=2x+-(x+-x-)=x++x.

(iii): By Proposition 3 and (i),

=

(x+

x-) 1\ 0 + x- = x 1\ 0 + x+ x- = -x- + x- = O.

V (-0))

We leave (iv)-(vi) to Exercise 3. (x + Ixl)/2 so using (vi), (vii): Note x+

I(x

+ IxD/2

(y + lyl)/21 = ~ I(x - y) + (Ixl- lyl)1

5.7. ORDERED LINEAR SPACES

199

t(lx+yl+(x+ y ))

(x+y)+

< 2"(l x l + x + Iyl + y) x+

+ y+.

(viii): From Proposition 3,

xVz

YVz=((x-z)VO+z)

(y

((y-z)VO+z)=(x-

z)+.

From (vii),

Ixvz

YVzl=j(x-z)+

(y

z)+IS:I(x-z)-(y

z)1

Ix

YI·

From Proposition 3 and (viii),

IxAz

YAzl=I-(-x)v(-z)+(-y)v(-z)IS:I(-y)

(-x)1

Ix-yl·

(ix) is left to Exercise 3. Remark 6 From (iv) [(ii)] and Remark 4, it follows that if X is an ordered vedor space such that x V 0 = x+ [Ix 11 exists [or x EX, then X is a Ricsz space. Example 7 F(8) is a Riesz space with

f

V g(t) = max{f(t),g(t)}, f A g(t) = min{f(t),g(t)}

for I, E 8. In particular, s is a Riesz space. Any vector subspace of F(8) which is a Riesz space under the inherited order from F( 8) is calJed a function space. For example, Coo, Co, C, [I, [00 are all function spaces. If 8 is a compad Hausdorff space, then C(8) is a function space; if Gk[O, 1] is the vedor space of all fundions 'f'; [0,1] -+ R which have k continuous derivatives, then Gk[O, 1] is an ordered vector subspace of G[O, 1] but is not a function space for k ;:: 1. Example 8 Let BV[a, b) be the vector subspace of F[a, b] consisting of the functions of bounded variation (A.I). BV[a, b) is a function space.

If X, Yare Riesz spaces and T ; X -+ Y is linear, then T is called a lattice homomorphism if T(x V y) = (Tx) V (Ty) for all x, y E X; if Tis 1-1, T is called a lattice isomorphism. For equivalent conditions, see Exercise 4. A linear map T ; X -+ Y between two ordered vector spaces is said to be positive if Tx ;:: 0 for every x ;:: o. A lattice homomorphism between Riesz spaces is positive. Theorem 9 Let X, Y be Riesz spaces and T : X -+ Y linear, 1-1, onto. 1fT and T-I are both positive, then T is a lattice isomorphism (onto Y).

TV

s:

s:

Proof: Let x, y E X. Since x x Vy, y x Vy and T is positive, Tx T(x Vy), T(x V y) so (Tx) V (TV) T(x V V). By symmetry, since T- 1 is positive,

s:

x V V = (T-J(Tx)) V (T-J(Ty))

s: T-1((Tx) V (Ty)).

200

CHAPTER 5. INTRODUCTION TO FUNCTIONAl, ANALYSIS

Hence, T(x Vy) ::;; (Tx) V (Ty). We next consider the order analogue of the dual of a NLS. If X is a. Riesz space, a subset Be X is order bounded if there exists y such that Ixl ::;; y for all x E B. If Y is another Riesz space, a linear map T : X -. Y is order bounded if T carries order bounded sets to order bounded sets and T is positive if T carries positive elements to positive elements. Let X- be the set of all order bounded (real-valued) linear functionals on X. Then X- is a vector space under pointwise addition and scalar multiplication and is ordered by the relation, f ::;; 9 if and only if f(x) ::;; g(x) for all x E X+. X· is called the order dual of X. It is easily checked that X· is an ordered vector space under t.his order and F. Riesz has shown that X- is actually a Riesz space. For this importallt result we first establish a lemma.

Lemma 10 (Riesz Decomposition) Let X be a Riesz space. If x, y, z E X+ and 0::;; z::;; x+y, then there exist Xl, YI E X such that 0 ::;; Xl::;; X, 0::;; YI ::;; y and Z=XI+YI' Proof: Set Xl = X 1\ Z, YI = Z Also) from Proposition 3,

Xl

Z

o ::;; YI'

O::;;YI

X 1\ z. Hence, 0 ::;;

z-xl\z=z+(-x)V(-z)

XI ::;;

X,

(z-x)VO::;;yVO

Xl

+ YI

Z)

y.

Ordered vector spaces which satisfy the conditions in Lemma 10 are said to have the decomposition property. Thus, a Riesz space has the decomposition property, buL there are ordered vector spaces with the decomposition property which are not RieBz spaces [see [P] p. 14].

Theorem 11 (F. Riesz) If X is a Riesz space, then X· is a Riesz space. Moreover, if f E X·, X E X+, then

(i) J+(x) (ii) r(x) (iii)

If I (x)

sup{f(y):O::;;y::;;x}, sup{-f(y):O::;;y::;;x}, sup{f(y):

Iyl ::;; x}.

Proof: From Remark 6, it suffices to show that f+ fVO exists for every f E X'. Let f E X· and for X E X+ set g(x) sup{f(y): 0::;; y ::;; x} [(i)]; note that since f is order bounded, g( x) is finite. Clearly

g(x) ~ 0, g(x) ~ f(x) and g(tx)

tg(x) for t ~ 0, X E X+.

(5.1)

We claim that g(x+y) = g(x)+g(y) for x, y E X+. If 0 ::;; Xl::;; X, 0::;; YI S; y, then XI+YI ::;; x+y so f(xI)+ f(yJ) = f(XI+YI)::;; g(x+y) andg(x) +g(y)::;; g(x+y). On the other hand, if 0 ::;; z ::;; x+y, by Lemma 10 there exist 0::;; Xl ::;; X, 0 S; Yl ::;; Y

0::;;

5.7. ORDERED LINEAR SPACES such that z

g(x

+ y)

=

+ YI. + g(y)

Xl

:::; g(x)

201

= f(XI + YI) = f(XI) + f(YI) :::; g(x) + g(y)

Then f(z)

so

For arbitrary X E X, X = x+ - x- so we can extend 9 to X by setting g(x) = g(x+) - g(x-). Note that if X = u - v with u 20, v 2 0, then x+ + v = x- + u so

or and the value of 9 does not depend upon how x is represented as the difference of two positive elements of X. From this and (1), it is easy to verify that 9 is a positive linear functional on X. We claim that 9 = j+ (= f V 0). If h E X- is positive with f :::; h, then f(y) :::; h(y) :::; h(x) for 0 :::; y :::; x so g(x) :::; h(x). From (1), 9 = j+. Formula (i) was established above. The remaining two formulas follow from (i) and f- = (-1)+ and If I = j+ +

r·

Corollary 12 If f E K and x E X, then If(x)1 :::; If I (Ixl). Proof: If I (Ixl) = sup{f(y) : Iyl :::; Ixl} 2 (f(x» V (f(-x». Since f V 9 = (f - g)+ + 9 [Proposition 5], for f, 9 E X- and x E X+ we have

(2) (f V g) (x) = sup{f(y)

+ g(x -

y) : 0 :::; y :::; x} and

(3) (f

+ g(x -

y): 0:::; y :::; x}.

1\

g)(x) = inf{f(y)

We have formulas "dual" to the formulas in Theorem 11. Theorem 13 Let X be a Riesz space, f E X

positive and x EX. Then

(i) f(x+) = sup{g(x) : 9 E K, 0:::; g:::; f}, (ii) f(x-) = sup{-g(x): 9 E X-, 0:::; g:::; f}, (iii) f(lxl)

= sup{g(x): 9 E K,

Igl :::; f}.

Proof: (i): If 0:::; 9 :::; f, then g(x) :::; g(x+) :::; f(x+) so

sup{g(x): 9 E X-,O:::; g:::; f}:::; f(x+). Conversely, define p : X -> R by p(u) = f(u+). Then p is a sublinear map on X [Proposition 5] such that p( u) 2 0 for all u EX. Let Y = {tx : t E R} and define h on Y by h(tx) = tf(x+). Then h is linear on Y and h(y) :::; p(y) for all y E Y. By the Hahn-Banach Theorem, h has a linear extension to X, still denoted by h, such that h(u) :::; p(u) for all u E X. If u 2 0, then h(u) :::; p(u) = f(u) and .

-h(u)

= h(-u):::;

p(-u)

= f((-u)+) =

f(O) = 0

'

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

202

so 0 :S h( u) :S f( u) for all u E X+. Hence, h E X- [Exercise 5], 0 :S h :S f, and

n·

f(x+) = h(x):S sup{g(x): 9 E X-,O:S g:S

Formulas (ii) and (iii) follow from f(x-) = f(( -x)+) and f (Ixl) = f(x+) + f(x-). We now consider NLS which have an order defined on them. A semi-norm (norm) on a Riesz space X is a lattice semi-norm (norm) if Ixl :S Iyl implies Ilxll :S IIYII. A semi-normed (normed) vector lattice is a Riesz space with a lattice semi-norm (norm); a complete normed vector lattice is called a Banach lattice.

Example 14 B(5), C(5), £=, c, Example 15

Coo

are Banach lattices.

Co

is a normed vector lattice which is not a Banach lattice.

Example 16 V (J!) is a complete, semi-normed vector lattice. For an example of a norm on a Riesz space which is not a lattice norIn see Exercise

7. Proposition 17 Let X be a semi-normcd vcctor lattice.

(i)

Illxlll

(ii)

Ilx+ - y+11

:S

Ilx - yll

so x

Illxl-lylll

:S

Ilx - yll

so x

(iii)

=

Ilxll, ->

->

x+ is uniformly continuous on X,

Ixl

is uniformly continuous on X.

Proof: (i) is easy. By Proposition 5, Ix+ - y+1 :S Ix - yl and Ilxl-lyll :S Ix - yl so (ii), (iii) follow. A linear subspace Y of a vector lattice is called a vector sublattice if x/\y, xVy E Y when x, y E Y. A linear subspace A of a vector lattice X is called an ideal (ordcr ideaQ if Ixl :S Iyl and yEA implies x E A. Since 1

x V y = 2(x

+ y + Ix -

yD,

every order ideal is a vector sublattice.

Example 18 B(5) is an ideal in F(5). C(5) is a vector sublattice of F(5) but is not an ideal. Similarly, Co and Coo are order ideals in £= but c is not. We now compare the order dual and the norm dual of a normed vector lattice.

Theorem 19 Let X be a normed vector lattice. Then X' is an order ideal in X- .

c

X- and, moreover, X'

5.7. ORDERED LINEAR SPACES

203

Proof: Let B c X be order bounded with Ixl ::; y for all x E B. Then Ilxll ::; Ilyll for all x E B so B is norm bounded. Hence, every x' E X' carries order bounded sets to bounded subsets and x' E X-. Assume x E X', x' E X' with lx-I::; Ix'i. We must show x E X'. Let x E X, IIxll ::; 1. Then for every y E X with Iyl ::; lxi, Ilyll ::; Ilxll ::; 1 so

Ix-(x)l::; lx-I (Ixl)::; Ix'i (Ixl)

SUp{X'(Y):

[Corollary 12 and Theorem 11]. Thus, x- E X' with

Iyl::; Ixl}::; Ilx'll

Ilx-1i ::; Ilx'll.

Remark 20 It follows from Theorem 19 that if x' E X', then x' is the difference of two positive, continuous linear functionals (x')+, (x't.

The argument above shows that

Ix'i ::; Iy'l in X'

implies

IIx'll ::; Ilylll.

Hence,

Theorem 21 The norm dual of a normed vector lattice is a Banach lattice.

The containment in Theorem 19 can be proper. Example 22 Define

f : Coo -+ R by f( {tj})

00

L: jtj

=

[finite sum]. Then

f is positive

j=1

and, hence, order bounded [Exercise 5]. But, f( e

j

)

= j so f is not continuous.

However, we do have

= X- .

Theorem 23 If X is a Banach lattice, X'

Proof: Suppose there exists f E X- such that exist Xk E X such that Ilx,,1I 1 and If(x,,)1 > F

Set y" =

t

"=1

Ix"l/k2.

f is not continuous. Then there

Then n+l'

IIYn+p

Ynll::;

E

1/k2

k=n+l so {Yk} is Cauchy in X and, therefore, converges to some Y E X. Clearly, 0 ::; Yn ::; Yn+h and we claim that Yn ::; y. First,

o ::;

(Yn

implies II(Yn - Y)+II ::; IIYn+p -

0::;

y)+::; (Yn+l'

yll

y)+::; IYn+1'

YI

for all n, p so

II(Yn - y)+11 ::;

Hence, (Yn - y)+ O. But, Yn Y::; (Yn Then Y 2 Yn 2. Ix,,1 /n 2 implies

li~IIYn+p - yll = o. y)+ = 0 implies 0 ::; Yn ::; Y as claimed. 2

2

Ifl(Y) 21f1(Yn) 2 If I (lx nl)/n 2If(x n )//n > n which gives the desired contradiction.

204

CHAPTER 5. INTRODUCTION TO FUNCTIONAL ANALYSIS

Theorem 24 The. completion of a normed vector lattice X is a Banach lattice. Proof: Recall the completion of X is C X" [Theorem 5.6.1.10] so by Exercise 8 it suffices to show that the canonical imbedding Jx J of X into X" preserves the lattice operations. For this it suffices to show that (J x)+ = J (x+) for x EX. Suppose x E X and f EX', f 2' O. We write Jx = x. From Theorems 11, 13 and 19, we have (x)+(f) = sup{x(g): 9 E X',O::::: g::::: f} sup{g(x): 9 E X',O::::: g::::: f} sup{g{x): 9 E K,O::::: g::::: f}

f(x+) That is, (Jx)+

(x+rU)·

J(x+) as desired.

Exercise 1. Show X+ is a convex cone, i.e., is convex and x E X+, t 2' 0 implies

tx E X+. Exercise 2. Complete the proof of Proposition 3. Exercise 3. Prove (iv)-(vi), (ix) of Proposition .5. Exercise 4. Let X, Y be Riesz spaces and T : X following are equivalent:

->

Y linear. Show that the

(i) T(x Vy) = (Tx) V (Ty), (ii) T(xI\Y)

(Tx) 1\ (Ty),

(iii) (Tx) 1\ (Ty) = 0 when x 1\ y

0 in X,

(iv) ITxl == T(lxl).

Exercise 5. If T : X -> Y is positive, show T is monotone in the sense that x ::::: y implies Tx ::::: Ty. Show every positive operator is order bounded. Exercise 6. If X is a semi-normed vector lattice show x uous.

->

x- is uniformly contin-

Exercise 7. Let BVo[a, b] be the linear subspace of BV[a, b] which consists of the functions which vanish at a. Show BVo[a, b] is a Riesz space under the pointwise order. Show the variation norm, Ilfll = Var'(f : [a, b]) is not a lattice norm on BVo[a, b]. [See Appendix 1 for the notation and definitions.]

5.7.

ORDERED LINEAR SPACES

205

Exercise 8. Show that the closure of a vector sublattice of a normed vector lattice is a vector sublattice. Exercise 9. If X is a normed vector lattice, show X+ is closed. Exercise 10. Let X be a normed vector lattice and {xd C X satisfy Ilxk - xll--+ 0, show x = SUp{Xk : k}. Hint: (x n - xm)+ = 0 for m > n.

Xk

:S

Xk+l'

If

Exercise 11. Let X be a normed vector lattice. Show x E X is positive if and only if I( x) 2': 0 for every positive, continuous linear functional I on X. Hint: Theorem 13.(ii). Exercise 12. Let X be a Banach lattice and x 2': 0, x E X. Show Ilxll

= sup{f(x)

: 0 :S IEX', IIIII

= I}.

Exercise 13. Let X, Y be normed vector lattices and T : X --+ Y positive. If X is a Banach lattice, show T is continuous. Hint: Theorem 23. Exercise 14. Show that any two complete lattice norms on a Riesz space must be equivalent. Exercise 15. Let X be an ordered vector space and Z a linear subspace of X such that whenever x E X there exists z E Z with z 2': x. Suppose F : Z --+ R is a positive linear functional. Show F has a positive linear extension to X. [This is a simple version of Kantorivich's Theorem; see [Vu] X.3.l.] [Hint: Set p(x)

= inf{F(z): z E Z,z 2': x}

and use the Hahn-Banach Theorem.] Exercise 16. Show there exists a bounded, finitely additive extension of Lebesgue measure on [0,1]. [Hint: Let X = B[O, 1],

Z = {f : [0, 1] --+ R :

I

is bounded and Lebesgue measurable}

and F: Z --+ R be F(f) = J~ Idm. Use Exercise 15.]

Cllapter 6 Ful1.ction Spaces 6.1

LP-Spaces

Let Jl be a measure on a a-algebra, L:, of subsets of S and let 0 < p < =. The space LP(Jl) consists of all real-valued L:-measurable functions, f, defined on S such that If IP is Jl-integrable. If Jl is Lebesgue measure on a measurable subset E of R n we write LP(Jl) =LP(E). If f, 9 E LP(Jl), then

If

+ glP

::; 2P(lf1 P + IgjP)

so LP(Jl) is a vector space under pointwise addition and scalar multiplication. Moreover, 0 ::; If I, 0 ::; f- ::; If I implies f- E LP(Jl) when f E LP(Jl) so LP(Jl) is also a vector lattice under the pointwise order. p For 1 ::; p < =, we set II flip = Us Ifl P dJl)1/ [the case 0 < p < 1 is covered in the exercises]. We show that 1IIIp is a semi-norm on LP(Jl)j note that 1IIIp is a lattice semi-norm. Only the triangle inequality needs to be checked. We do this with the aid of two preliminary results.

r :;

Lemma 1 Let a, b:2: 0 and 0

r,

< t < 1. Then at b1- t

::;

ta + (1 - t)b.

Proof: Define ip: (0,=) --+ R by ip(s) = ts - st. Then ip'(S) < 0 for 0 < s < 1 and ip' (s) > 0 for s > 1 so ip has a minimum at s = 1. Thus, t - 1 ::; ts - st for s > O. Put s = alb [if b = 0, trivial] and multiply by b to obtain the desired inequality. Theorem 2 (Holder's Inequality) Let 1 < P < = and q be such that lip + II q = 1. If f E LP(Jl), 9 E U(Jl), then fg E L 1(Jl) and IIfgl11 ::; Ilfllp Ilgllq· Proof: If either Ilfllp = 0 or Ilgllq = 0, the result is trivial so assume that both are positive. Note that if we set t = lip in Lemma 1, the conclusion reads AB::; Nip + Bqlq when A, B:2: O. Set A = If(t)1 I II flip' B = Ig(t)1 I Ilgllq for t E S to obtain .;.;:.If~(t..c.::;)g.,.:..(tfc'-)I < _If_(t)_IP + _lg_(t)_lq II flip Ilgllq - p Ilfll~ q Ilgll: 207

CHAPTER 6. FUNCTION SPACES

208

IIfglll IIfllp Ilgllq Remark 3 For p Inequality.

1 p

1 +-=1. q

2, this inequality is often referred to as the Cauchy-Schwarz

We can now easily obtain the triangle inequality for 1111p.

Theorem 4 (Minkowski Inequality)

Iff, 9 E LP(p),

1 ::; p

<

00,

then

Ilf + gllp ::;

IIfllp + Ilgllp· Proof: For p

1, the inequality is clear. Assume p > 1. Note

Is If + glP dp = Is If + gllf + glP-l dp ::: Is If I I! + glP-1 dp + Is Igllf + glP-1 dp.

(6.1 )

Observe that (p l)q = p and apply Holder's Inequality to both terms on the right hand side of (1) to obtain

Is If + glP dJL ::: Us Ifl P dJL// pUs(lf + glP dJL //q + Us IglP dl')l/P Us If + glP dJL)I/q. If

Is If + glP dJL = 0, the result

(6.2)

is trivial; otherwise, divide (2) by

(is If + glP dJL) I/q to obtain the desired inequality. We now extend the Riesz-Fischer Theorem to p

Theorem 5 (Riesz-Fischer) For 1 ::; p < Proof: Let

!k

00,

>

1.

LP(p) is complete under II lip .

be an absolutely convergent series in LP(JL). Define gn(t) =

if.(t)1 for t E S. Then n

Ilgnllp Now 9n i so if we set g(t) 9n 2: 0, so by the MCT

00

Ilfkllp ::; I: IIfkllp '=1

= limgn(t),

M

<

00.

then 9 is L:-measurable and gn(t)"

i g(t)P,

209

6.1. LP-SPACES

Hence, 9 is finite fi-a.e. and we may assume that 9 E Lp(fi}. Note that this means <Xl

that the series 2: If,,(t)1 converges in R for fi-almost all t E 5.

"=1

fk (t) when this series converges in Rand f (t)

Define f by f(t)

0 otherwise.

Then f is 2:-measurable and f is the fi-a.e. limit of 2:f". Let Sn

fIe. Then

ISnl : : : gn ::::: 9 so If I : : : 9 fi-a.e. and f E Lp(fi). Also, ISn - flP ::::: 2 and Sn -. f fi-a.e. so by the DCT, lis" flip -> O. By Theorem 5.1.4, Lp(fi) is complete. Note that IIfll1' 0 if and only if f = 0 fi-a.e. so if Kp = {J E LP(fi) . Ilfllp OJ, then £1'(fi)= L1'(fi)1 Kp is a Banach space and two functions are in the same coset of O(fi) if and only if they are equal fi-a.e. (§5.4). Instead of working with the coaets in D'(fi), it is customary to identify two functions if they are equal almost everywhere P

and treat the cosets as if they were functions. Under this agreement, we would say that O(fi) is a Banach lattice. Comparison of LP-spaces: In general, there are no inclusion results for LP-spaces. For example, if f(t) 11 Vt for 0 < t ::::: 1 and f(t) 0 otherwise, then f E Ll(m) but f t/: L2(m), while if g(t) = lit for t 2: 1 and g(t) 0 otherwise, then 9 E L2(m) but 9 t/: Ll(m). [See also Exercises 1 and 2.] For finite measures, we do have Proposition 6 Let fi be a finite measure and 1 ::::: r < S <

00.

Then L3(fi) C LT(fi)

and the inclusion map is continuous. Proof: Let h E L8(fi). Set p obtain

sir, f =

IhlT

and 9

1 in Holder's Inequality to

is Ihlr dfi ::::: (is Ihl') Tis (fi(5))1-r/8.

Apply Theorem 5.2.1 to obtain the continnity. If fi is counting measure on 5, we set £1'(5)= L1'(fi), and if 5 £1'(N) =p. Thus, £1' consists of all sequences {tj} such that

N, we write

[recall Example 5.1. 7]. For £1'-spaces, in contrast to Proposition 6, we have

< s < 00, I'" C £8 and the inclusion map is continuous. Moreover, the containment is proper.

Proposition 7 If 1 ::::: r

Proof: Suppose x {tj} E I'" with Ilxlir : : : 1. Then Itj I : : : 1 for all j so Itj IS : : : Itj IT which implies IIxll: : : : Ilxll;. If x {tj} E i!: with x '" 0, then x/llxli r E i!: and Ilx/llxllrllr 1 so by the observation above IIx/llxllrll. : : : 1 or IIxll. : : : IIxlir.

CHAPTER 6. FUNCTION SPACES

210

Take tj = (1/j)1/T. Then x = {tj} E fls\fl!. For a thorough discussion of the possible inclusion results for LP-spaces see Romero

([RD· Dense subsets of LP : We have the analogues of Theorems 2, 3 and 6 of §3.5 for LP-spaccs [Exercise 6].

Theorem 8 The vector space of L,-simple functions in Y(/1) is dense in Y(/1). Moreover, given f E Y(/1) there exists a sequence of simple functions {'Pd in Y(/1) such that 'Pk --+ f pointwise and II'Pk - flip --+ 0; if f 2 0, the {'Pd can be chosen such that 'Pk f·

r

Assume that /1 is a premeasure on the semi-ring S of subsets of Sand tbat L, is the a-algebra of /1* -measurable subsets of 5 (§2.4). Let /1 denote the restriction of /1: to L,. Theorem 9 The vector space of S -simple functions in Y(/1) is dense in Y(p). Theorem 10 Let 5 be a locally compact Hausdorff space and /1 a regular Borel measure on B(5). Then C c (5) is dense in L P (/1).

As an application of Theorem lO we give a generalization of Theorem 3.11.10 to LP-spaces. Theorem 11 If f E Y(Rn), then lim Ilfh - fll = h-+O

P

o.

Proof: Let E > O. By Theorem 10, there exists 'P E Cc(Rn) such that Ilf - 'Pllp < 'P is uniformly continuous and there exists a such that 'P(x) = 0 for Ilxll 2 a so there exists 0 < 8 < 1 such that 1'P(x + h) - 'P(x)1 < E for Ilhll < 8. Thus, if Ilhll < 8, Eo

then

(6.3) Now if

Ilhll < 8, Ilfh -

flip

~ Ilfh -

'Philp + II'Ph - 'Pllp + II'P - flip < 2E + II'Ph - 'Pllp

so the result follows from (3). Convergence in LP : We give a characterization of sequential convergence in LP for 1 eralizing Proposition 3.7.1, we have Proposition 12 If /k

--+

f in Y(/1), then

/k

--+

~

p <

00.

Gen-

f /1-measure.

We give some additional necessary conditions that must be satisfied by a sequence converging in LP.

6.1. LP-SPACES

211

Definition 13 Let F be a family of signed measures on L:. continuous if lim veE) = 0 uniformly for v E F.

F is uniformly p-

I'(E)-O

Proposition 14 Let fk E Y(p) and vk(E) :::: IE 1/kIP dp for EEL:, /k -+ fo in LP(p), then {VI<; : k ;::: O} is uniformly p-continuous.

k=

0,1, .... If

Proof: Let ( > O. 3 N such that k ;::: N implies IIh - follp < t.. By 3.2.17 there exists" > 0 such that E E peE) < 6 implies vJ,;(E) < ( for k = 0,1, ... , N. For k ;::: Nand peE) < 6,

vk(E)I/P

~

(i If;, -

fol Pdp) IIp +

(i Ifol Pdp) IIp ~ IIh - follp + VO(E)I/P < (+ (lIp.

Definition 15 Let F be a family of signed measures on L:. F is equicontinuous from above at 0 if E" E E" ! 0 implies lim v( E/;) = 0 uniformly for v E :F. k

Proposition 16 Let /k E LP(p) and vJ,;(E) = IE 1/kIP dp for EEL:, k fk -+ fo in Y(p), then {Vk : k ;::: O} is equicontinuous from above at 0.

= 0,1, ....

If

Proof: Let f > O. There exists N such that n ;::: N implies Ilfn - follp < f. Let E" E L:, E" ! 0. By 3.2.9 and there exists K such that Vn(Ek) < f for k ;::: 1< and n = 0,1, ... , N - 1. For n ;::: Nand k;::: K,

Vn(E,,)l/p ::;

(i. Ifn

folP dp) IIp

+

(i. Ifol Pdp) IIp <

f

+ flIp.

Propositions 12, 14 and 16 give three necessary conditions for a sequence to converge in Y. It is an interesting result of Vitali that these necessary conditions are also sufficient. Theorem 17 (Vitali) Let 11, E Y(p.) and vk(E) = k = 0,1, .... Then I" -+ fo in Y(p.) if and only il

(i) /k

-+

IE If"I Pdp.

for EEL

fo p-measure,

(ii) {Vk: k ;::: O} is uniformly p.-continuous, (iii) {Vk: k ;::: O} is equicontinuous from above at 0. Proof: => follows from the above. -¢::: Since {t: !k(t) =f O} is p o--finite [3.2.14J, we may as well assume that p. is k=O o--finite. Let S Ek where Ek E L:, E" i and P.(Ek) < (Xl. Set Fk = EA:. k=1 Let f > O. From (iii) 3k such that IF. II; IP dp. < (E/2)p for j ;::: O. Hence, for j ;::: 0,

U

U

(L.lfi

I;I Pdp. rIP::; (L.lliIP dp. rIP

+ (L.II;I Pdp. rIP < (

(6.4)

CHAPTER 6. FUNCTION SPACES

212 Let G;j = {t : If;(t) - fj(t)1 ~ t}. Then

(6.5) ::::: IEknG;j If; - fjlP dJl

+ tPJl(Ek).

From (i) and (ii) the first term on the right hand side of (5) goes to 0 as i, j --+ 00. From (4) and (5), it follows that {t;} is a Cauchy sequence in LP-norm. By the RieszFischer Theorem there exists 9 E LP(Jl) such that f; --+ 9 in LP(Jl). By Proposition 12 f; --+ 9 in Jl-measure so fo = 9 Jl-a.e. and f; --+ fo in LP(Jl).

Dual of LP for p > 1 : We now give a characterization of the dual of LP(Jl) for 1 < p < follows p is fixed, 1 < p < 00, and l/p + l/q = 1. Proposition 18 If f E Lq(Jl), then F,: LP(Jl) --+ R, defined by F,(g) a continuous linear functional on LP(Jl) with IIF, II = Ilfll q.

00.

=

In what

IsfgdJl, is

Proof: By Holder's Inequality, 1F,(g) I : : : Ilfllq Ilgllp so F, is continuous with IIfll q· q q Set 9 = Ifl - 1 sign f. Then 9 is measurable and IglP = Ifl(q-l)P = Ifl so 9 E LP(Jl) and q q q F,(g) = Is Ifl dJl = (Is Ifl dJl)I/q (Is Ifl dJl)I/P = IIfllq (Is IglP dJl)I/P = IIfllq IIgllp·

IIF,II : : :

Hence, IIF,II = IIfll q· Thus, the map U : Lq(Jl) --+ LP(Jl)', Uf = F" is a linear isometry which is order preserving. We show that U is onto, i.e., every continuous linear functional on LP(Jl) has the form F, for some f E U(Jl). Hence, we can identify U(Jl) and LP(Jl)' as Banach lattices. We first establish a useful lemma, sometimes called the Reverse Holder Inequality. Lemma 19 Let Jl be a-finite and let f : S

M = sup

--+

R be measurable. Suppose

{is Ifgl dJl : IIglip : : : I} <

00.

Then f E Lq(Jl) and IIfllq = M. Proof: Let S =

UAk with Ak 1 Sand Jl(A k) <

00

and

k=l

Ek = {t E Ak : If(t)1 ::::: k}.

Ifl q- 1 GEk sign f.

Set gk = Then gk is bounded, measurable and vanishes outside Ak q so gk E LP(Jl). Moreover, gd = Ifl GEk so

6.1. £P -SPACES

213

and

Ifl q CE.d!J,) I-l/p=l/q ~

(is

q

M.

q

Since Ifl C E • 1 I/l , the MCT implies I E U(p,) with 1I/11q ~ M. The reverse inequality M ~ 1I/IIq now follows from Holder's Inequality. See also Exercise 14.

Theorem 20 (Riesz Representation Theorem) If F E LP(p)', then there exists IE Lq(p) such that F(g) Is Igdp for 9 E LP(p). Moreover, IIFII = 1I/11q· Proof: First assume that fl is finite. Define v : --+ R by vCE) = F(CE) [note CE E LP(fl) since p, is finite]. Then v(0) F(O) = 0 and v is finitely additive since

F is linear. We claim that v is actually countably additive. For if {Ej pairwise disjoint and E

UE

j ,

}

c

L: are

then

j=l

p

so vee) -

t

v(Ej )

--+

O. Since

j=l

v is a finite signed measure which is absolutely continuous with respect to fl.

By the Radon-Nikodym Theorem, there exists a fl-integrable function f such that = Fe CE ) for every E E By linearity, we have F( 'P) = Is l'Pdp for every L:-simple function 'P. We claim that F(g) = IsIgd/l for every 9 E LP(p,). For this we lJlay assume 9 "2': o. Let A {t : I(t) "2': OJ, B = {t : f(t) < OJ. Choose a sequence of non-negative, simple functions {'Pk} such that 'Pk 1 9 and II'Pk - gllp ....... 0 [Theorem 8]. Then 'PkCAI 'PkJ+ 1 gJ+ and II'PkC A - gCAIl p --> 0 so by the continuity of F,

veE) = IE Idp,

F('PkCA)

L

f'Pkdp

isJ+'Pkdfl

-->

F(gCA)

and by the MCT gJ+ E U(p,) and

L

l'Pkdfl

is r 'Pkdfl 1 is r gdp.

Hence, F(gC A) IsgJ+dp,. Similarly, F(gCB) Isgl-dfl so F(g) = IsIgdp,. That IE Lq(fl) and IIFII Ilfllq follows from Lemma 19. Now let p, be an arbitrary measure. For E E 2:, let

CllAPTER 6. FUNCTION SPACES

214

{E E L : p(E) < oo}. By the part above, for every E E £ there exists fE E Lq(E) [unique up to p-a.e.] such that F(CEg) = IsiEgdp for 9 E LP(p).

Set £

Moreover,

IIfEllq = sup {IF(CEg)1 : IIgllp::;

I}::; I!FII·

!B p-a.e. Let a = sup{lIfEll q : E E £} ::; I!FII. If A, B E £ and A C B, then fA in S so !fA 1 ::; IfBlp-a.e. and IIbll q ::; IIfBll q • Hence, there exists an increasing sequence {Ek} C £ such that IIfE.ll q i a. Let f (t) = (t). Since fHl !k p-a.e. in E k, limfE. (t) exists for p-almost all t E S, f = fk p-a.e. in Ek and f 0 off E

UEk.

k=]

We claim that F vanishes on LP(EC). For if F is not zero on LP(EC), then since the simple functions are dense in LP(p), there exists B E £ such that B C Ee and F does not vanish on LP(B). Therefore, fB f. 0 p-a.e., and since B n Ek 0, q

a ~ IlfBuE.II: IIfBII: + lliE.lI: which implies aq ~ IIfBII: + a9 and gives the desired contradiction. Let 9 E LP(p). Since f E L9(/-L), fg E l}(p). Since giE. --t gf /-L-a.e. and IgiE.1 ::; Igfl, the DCT implies IsgiE.dp --t Isgfdp. Similarly,lIgCE• gCElip --t 0 so the continuity of F implies F(gCE.) --t F(gCE). Hence, limF(gCE.)

lim hgiE.dp

Corollary 21 If 1 < p <

00,

hfgd/-L == F(CEg)

F(CEg

+ CEeg)

F(g).

then P(p) is reflexive.

We describe the dual of certain Ll-spaces in §6.2. We give two interesting applications of duality. The first is an extension of Minkowski's Inequality to integrals. Minkowski's Inequality asserts that the D'-norm of a sum is less than or equal to the sum of the LP-norms. We give a generalization where the sum is replaced by an integral. Theorem 22 (Minkowski's Inequality for Integrals). Let (S,S,/-L), (T, T, v) be ufinite measure spaces and f : S x T --t R measurable with respect to the u-algebra of

(p X v)*-measurable sets. Let 1 ::; p < 00 and assume that f(·, t) E LP(/-L) for t E T and t --t IIf(', t)lIp belongs to V(II). Then f(05,·) E Ll(lI) for p-almost all s E S, the function s --t IT f( s, . )dll belongs to LP(p) and

Proof: The case p 1 follows from Fubini's Theorem. Let 1 hE L9(p). By Fubini's Theorem and Holder's Inequality,

Is {IT If( s, t)1 dll(t)} Ih(s) I dp( s)

<

p

<

00

IT Is If(s,t)llh(05)1 dp(s)dll(t)

< IT Us If(s, tW dp(s)/Ip IIhllq dll(t) IT IIf(·, t)lIp IIhllqdll(t).

and

6.1. £P -SPACES

By Lemma 19, s

-->

215

JTf(s,t)dv(t) belongs to LP(p) with

Application to Convolution: Recall the convolution product of two functions

f*g(x)=

r

JRn

f,

9 :

Rn

-->

R is defined

to be

then f * 9 E

LP(Rn)

f(x-y)g(y)dy

[§3.11]. Proposition 23 Let 1

and

Ilf * gllp

:0::::


If f E

Ll(Rn),

1 and h E

U(Rn).

CXl.

9 E

LP(Rn),

II fill Ilgllp·

Proof: Let l/p Fubini's Theorem,

+ l/q =

JRn JRn If(x - y)g(y)h(x)1 dydx

Then, using Exercise 3.11.8 and

JRn Ih(x)1 JRn If(x - y)g(y)1 dydx JRn Ih(x)1 JRn If(t)g(x - t)1 dtdx JRn If(t)IJRn Ih(x)g(x - t)ldxdt < JRn If(tJlllg-tllpIlhllq dt JRn If(t)lllgllp Ilhllqdt

(6.6)

II fill Ilgllp Ilhllq. Since h can be taken to be non-zero everywhere [e-lIxI12j, (6) shows that f * 9 is finite a.e. The inequality in (6) shows that f * 9 induces a continuous linear functional on U (Rn). By Theorem 20, f * 9 E LP (Rn) and by (6), IIf * gllp :0:::: Ilflll Ilgllp. We can combine Theorem 22 and Proposition 23 to obtain an extension of Theorem 3.11.11 to the case where 1 < p < CXl.

{ipd be Ilf * ipk - flip --> o.

Theorem 24 Let

Then

an approximate identity and f E

LP(Rn),

Proof: f * ipk E LP(Rn) by Proposition 23. Denote the function t fx. By Minkowski's Inequality for Integrals,

1

-->

:0::::

f(x

<

CXl.

+ t)

by

P

The function g(y) = Ilf-y - flip is bounded, continuous [Theorem 11] and g(O) = 0 so the right hand side of (7) converges to 0 as k --> CXl by Exercise 3.11. 7. Exercise 1. Let S = (0,1/2]' 1 but not in L'(S) for r > p.

:0::::

P<

CXl.

Show f(t)

=r

l P /

(ln(1/t))-2/ p is in LP(S)

CHAPTER 6. FUNCTION SPACES

216

Exercise 2. Let S [0,00), 1 :5 p < 00. Show f(t) "" t- 1 / 2 (1 but not in LP(S) for p '" 2. Exercise 3. If fk fin LP(Il) and gk -+ gin U(Il) for 1 show that fkgk -+ f 9 in £1 (11)· Exercise 4. If

f

E £'(11) n L"(Il), 1 < r <

S

+ In Itlt1

is in £2(S)

< p < 00, l/p + l/q

= 1,

< 00, show f E LP(Il) for r :5 p :5 s.

Exercise 5. Let gn, 9 be measurable functions such that Ignl :5 M and gn -+ 9 Il-a.e. show that if fn -+ fin LP(Il), then fngn -+ fg in V(Il). Exercise 6. Prove Theorems 8-10. Exercise 7. Show the polynomials are dense in VIa, b) for 1 :5 p to Rn.

< 00. Generalize

Exercise 8. Let l/p+l/q 1, f E V(R), 9 E Lq(R). Show F(t) is uniformly continuous on R.

fR f(x+t)g(x)dx

Exercise 9. J.
If - glp

Exercise 10. If fk -+ fin LP(Il) for some 1 :5 p < 00, show there exists a subsequence of {fA:} which converges Il-a.e. to f. Exercise 11. Let fA:, f, 9 E LP(Il) and fk -+ f Il-measure. If Ifkl :5 9 Il-a.e., show fA: -+ f in V(Il) [Dominated Convergence Theorem). Exercise 12. Let 1 < p < 00, l/p+l/q = 1. Suppose {fJ<} c LP(Il) and limfs /kgdll exists for every 9 E Lq(Il)· Show there exists f E V(Il) such that lim Is fkgdll =

Is fgd~l. Exercise 13. Let fk, fo E L2[a, b] and Ilfk - fol1 2 Fic -+ Fo uniformly on [a, b). Is p "" 2 important?

-+

O. Set Fk(t) "" I~ k

Show

Exercise 14. Let 11 be O"-finite and 1 < p < 00. Suppose f is measurable and fg E Ll (11) for every 9 E Lq (11). Show f E V (11). Show O"-finiteness cannot be dropped. [Hint: Exercise 2.2.16.J

6.1. LP-SPACES

217

Exercise 15. If p, is finite, show condition (iii) of Theorem 17 can be dropped. Exercise 16. Let k : S

X

S -+ R be in L2 (p,

X

p,). Show

(i) y(.s) = Is k (.s, t) x (t) dp, (t) exists for p,-almost all s E S when x E P (p,) and (ii) yEP (p,) with

IIYl12 .::; IIk11211x112'

The map K : x -+ y is a continuous linear map from L2 (p,) into itself and is called an integral operator. The function k is called the kernel of K.

Exercise 17. Show Exercise 18. Let A

f1'

(S), 1 .::; p

< 00, is separable if and only if S is countable.

[a,j] be an infinite matrix and suppose A maps iT into £! in

the sense that the sequence Ax

= L~l a,j x j }, E £! for every x =

CGT to show that A is continuous.

{x j} E

fT.

Use the

218

CHAPTER 6.

6.2

FUNCTiON SPACES

The Space Loo(l1)

Let 11 be i1 measure on it CT-algebra L:: of subsets of S. The spa.ce 1/''''(/L) consists of all L::-measurable functions f: - t R such that there exists a l1-null set N with f bounded on S\IV. is a vector space under pointwise addition and scalar multiplication and is a vector lattice under t.be pointwise order. We define a seminorm on I:"'(I1) by

IIIII00

II(I)I : I E

illf

S\IV} : It(N) = O}.

(6.1)

Thc functions in a.re called It-essenlially bounded Iundions and the semi-norm in (I) is ca.lled the 11- essential .'mp o[ f. If F' c R" is Lebesgu(, mcasurable anrl m is lIwasure Oil E, we write L The following properties should be clear.

Proposition 1 (i)

f =0

o if and

III ".

only if

/f-a.f.

Ilflla:. (iii) II III

0: l1{t:

(ii)

S;

Igl

thm

It-a.

IfU)1

IIJlcv

M} '" O}.

S;

so II

lie<. is

Il

{attire norm on LO-C(p).

As was the ca.se for Uli' ot.her LP-spacco, Wf' a.gree to idcntity f1!llctions in [/''''(/f) which are equal /f-a.('. so we regard L "'(/I) as a NLS. The ana.logue of the RieszFischer Theorelll holds for L CY.' (p).

Theorem 2 Proof: Let

~s

complete

allriPl'

Uk 1 be Cauchy ill 1/>0(//.)'

Thell there exists a li.-nul! set lV such that [or l

tt

IV.

(6.2)

is Cauchy for t S\1"/; set lim fdl) for t E S\I\' and f is lIl('as\II'ablc. Lel ( There exists 1'1 such that k, j - t 00 in (2) impli(~s that J /:1 < L Fixing k ~ n a.nd II.ik . . . fll.X) S; E so I and .h - t f in L=(jl). Thus, LOO(I1) is a. Banach lattice. We now show that the dual of Ll (Il) is [or CT-linite Ii. From Exercise 1, we have Therefore, f(t) = 0 [or t

N. Then

Proposition 3 Lei f E L=(ll). Define F f : V(I!.) F f is a continuous linear functional on Ll(p) with

-t

R by Ff(g)

IIFfl1

=

Is Igdp.

Then

Ilfll oo '

For CT-finite measures we show t.hat every continuous linear functiona.l on [,1(,1.) has the form Ff for some f L OO(p).

6.2. THE SPACE LCC(p.)

219

Theorem 4 (Riesz Representation Theorem) Let p. be a-finite.

there exists I E LOO(Ji) such that F(g) is unique up to p.-a.e.J.

11 FE (V(Ji»', 11/11,,,, !I

Is Igdp. lor g E LI(Ji) and IIFII

Proof: First assume that p. is finite. Then as in the proof of Theorem 6.1.20 there exists I E LI (p.) such that F(g) = Is I gdp. for g E LI (p.). We claim that IE Loo(p.) and IIFII = 11/1100' Let f > O. Set

= {t:

IIFII

+ 1')p.(A) :s;

L

A Then (l1F11

+ I' <

III dp.

I(t)}.

L

I(sign f)dp.

F(CA sign f) :s; IIFIIIICA sign IIII = IIFII p.(A).

=

Hence, p.(A) 0 and I E LOO(p.) with 11/1100 :s; IIFII + f so 11/11"" :s; IIFII. By Proposition 3, IIFII :s; 11/1100 so 11/11"" = IIFII· Assume that p. is a-finite. Let {Ek} be an increasing sequence from I: with Ek 1 Sand P.(Ek) < 00. By the first part, there exists a sequence {/d c LOO(p.) such that Ik vanishes outside E k , IIlkll oo :s; IIFII, Ik = Ik+l in Ek and F(g) Is Ikgdp. for every 9 E Ll(p.) which vanishes outside Let I = limlk so I Ik in Ek and 11/1100 S I/ F II. Therefore, F(CE.g) IsI(CE.g)dp. for every 9 E V(p.). But CE.g --+ 9 pointwise and ICE.gl :s; Igl so by the DCT and the continuity of F,

F(g)

= limF(CE.g) = lim

is I(CE.g)dJi = is IgdJi.

From Proposition 3, IIFII :s; 11/1100 so IIFII 11/11",,· Thus, when p. is a-finite the map U ; Loo(Ji) --+ (V(Ji»', U(f) = Ff is a linear isometry from L=(Ji) onto (Ll(p.»)' which is obviously order preserving. As before, we write LOO(p.) = (Ll(p.»'. In contrast to the situation when 1 < p < 00, Theorem 4 is false if the a-finiteness condition is dropped. Example 5 Let S = R and let I: be the <1-algebra which consists of the sets which are countable or have countable complements. Let p. be counting measure on Then Ll(p.) consists of the functions 9 on R which vanish outside a countable set and satisfy Ilglll I: Ig(t)1 < 00. Define F : LI(p.) --+ R by F(g) = I: get). Then tER

1>0

FE (Ll(p.»' and IIFII = 1. If I: S

F(g) = then

1= C(o.",,).

--+

is Igdp.

R were to satisfy

Lg(t) for 9 E Ll(p.), t>o

But this function is not I:-measurable. [See, however, Exercise 5].

Remark 6 There is a description of Ll(p.)' for arbitrary measures due to J. Schwartz ([Sc]). There are necessary and sufficient conditions known for the measure p. to satisfy Ll (p.)' = Loo (p.); see [TT].

CHAPTER 6. FUNCTION SPACES

220

Recall that for 1 < p < 00, LP(JL) is reflexive [6.1.21]. However, U(JL) is not generally reflexive, even for finite measures.

Example 7 Let S = [0,1] with Lebesgue measure. Then C[O, 1] is a closed subspace of L=[O, 1] (Exercise 2). Define F : C[O, 1] ---> R by F(:.p) ',0(0). TheIl F is a continuous linear functional on C[O, 1] with IIFII = 1. By the Hahn~Banach Theorem, F can be extended to a continuous linear functional F: LOO[a, 1] R with IIFII = 1. Now there exists no f E LI[O, 1] such that P(g) Id fgdm for all g 1/'°[0, I]. For, choose continuous gk such that Igk(t)1 ::; 1 for 0:::: t ::; 1, gk(O) = 1 and gk(t) ---> 0 for 0 < t ::; 1. Thus, if such an f exists, by the DCT limId fgkdm = 0 while F(gk) = 1 for each k. See also Exercise 3 for another proof. It follows from Exercises 3 and 5.6.1.1 that LOO[O, 1] is not reflexive. We descrihe the dual of L= in §6.3. We establish a result which in sOrrle sense justifies tbe notation for

Proposition 8 Let J1. be a finite measure. If f E L=(J1.), then Proof: If 11/11= 0, the result is trivial so assume Set H = {t: I/(t)1 ;:: k}. Then J1.(Ji) > 0 and

Ilfll oo '

kP JL(H) ::;

LIII

P

dJ1. ::;

Ilfll oo

IJiloo >

= lim

p~oo

Ilfll p .

0 and let () < k <

[I/I P dJL ::; lifll;:' IL(8)

so

Letting p

---> 00

gives k ::;

IIfllp ::; 11/1100'

Hence, the equality follows.

If f L=(JL), then Ig E LP(J1.) for every g E LP(JL). As an application of the CGT, we establish the converse of this result. Example 9 Let f : S ---> R be measurable and suppose Ig E U(tl) for every g LP(J1.). Then I E LOO(JL). Define a linear map T : U(JL) ---> U(p) by Tg = fg. We Igk~' h in U(J1.). Then t.here is claim that T is closed. Suppose gk ---> g and Tg k a subsequence {gn.} converging J1.~a.e. to g so fgn. fg = Tg Il~a.e. [6.1.12,3.6.3]. hand T is closed. By the CGT, T is continuous, ami we may assume Hence, Tg IITII ::; 1. Let 8> 0 and set E {t: If(t)1 2: 1 + 8}. Since 111'ngll ::; IIgll for n ;:: 1,

[lgl P dJ1. ;:: and since (1 + 8)np ---> 00, IE f E LOO(J1.) with IIfll oo ::; 1.

[WgI P•dJ1.

;::

Igll' dJ1. = 0 for

1:

(1

+ 8tP IglP dJ1,

every g E LP(JL) so p(E) = O. lIenee,

6.2.

THE SPACE LOO(p)

221

Integrating Vector-Valued Functions We indicate how the CGT and the duality between LI and Leo can be used to define an integral, called the Gelfand integral, for vector-valued functions. Let X be a Banach space and p a measure on the u-algebra r:; of subsets of a set S. Let I: 5 --+ X be such that x' 0 I = xiI is p-integrable for every x' E X'. The mapping F : X' --+ £I (p) defined by F (x') = xiI is linear and has a closed graph (Exer. 10). Hence, F is continuous by the CGT. For E E r:;,

x'

--+

kXlldp

= (CE,F(x'))

defines a continuous linear functional on X' since I(CE,F(x'»)1 ~ 11F1l11x'll. This continuous linear functional is called the Gelfand integral of lover E and is denoted by IEldp. Thus IEldp E X" and UEld{t,x') = IEx'ldp. If IEldp EX JxX) for every E E r:;, then I is said to be Pettis integrable and IE Idp is called the Pettis integral of F over r:;. Of course, if X is reflexive, every Gelfand integrable function is Pettis integrable, but, in general, this is not the case.

Example 10 Let It be counting measure on N and define I: N --+ Co{) by I(k) Let x' = {tk} E II (eo)'. Then Xl I (k) tk so x'l is p-integrable and

kx'ld{. Hence, IEldp integrable.

= CE and

L

tk =

k 10 •

(CE,x').

kEE

when E is infinite IEldp

= CE E

lOO\eo so I is not Pettis

For more information on the Gelfand and Pettis integrals, see [Du]. There is also a dose analogue of the Lebesgue integral for vector-valued functions, called the Bochner integral; this integral is also discussed in [Dul.

Exercise 1 (Holder's Inequality for p = 1). If IE LCO(p), 9 E LI(p), show Ig E LI(p) and Il/gllI ~ 11111"" IIglll' Exercise 2. If 'P E C[O, 1], show sup{l/(t)1 : 0 ~ t ~ I} equals the m-essential sup of 'P. Is this equality valid for an arbitrary measure on [0,1]7 Exercise 3. Show £I [0,1] is separable and L"" [0,1] is not separable. Use Corollary 5.6.1.13 to show that £1[0,1] is not reflexive. Exercise 4. Repeat Exercise 3 for II and lOO.

222

CHAPTER 6. FUNCTION SPACES

Exercise 5. Let S '" 0. Let [I(S) LI(p.), where p. is counting measure on S, and [oo(S) Loo(p.) [so [OO(S) consists of all bounded functions on S]. Show [I(S), = [oo(S). Exercise 6. Let 1 5 p < 00 and lip + llq = 1. Show that if I E £p(R), 9 E Lq(R), then 1* 9 is defined everywhere, is uniformly continuous, and II! * giL"" 5 III lip Ilgllq· Exercise 7. Let p. be (i-finite. Show 9 E Loo(p.) satisfies 0 5 9 5 1 p.-a.e. if and only if 05 Is Igdp. 5 Is Idp. for every I E LI(p.) with I 2 O. Exercise 8. p. is non-atomic if every EEL with p.(E) > 0 contains a subset A E L with 0 < p.(A) < p.{E) [i.e., Lebesgue measure]. If p. is finite and non-atomic, show L""{p.) is infinite dimensionaL Can the non-atomic assumption be dropped? Exercise 9. If I E L""'(p.) n J}{p.), show I E £P(p.), 1 < p < Exercise 10. Show the operator F: X' a closed graph.

->

00.

J} (p.) defined in the last paragraph has

Exercise 11. Let p. be counting measure on N and I : N -> Co be defined by I (k) = ek I k. Find the Gelfand integral of I with respect to p.. Is I Pettis integrable? Exercise 12. Let

{In} such that

I

E L= [a, bJ. Show there exists a sequence of continuous functions

IIlnll"",

5

11/11= and limJ.b Ingdm n a

J.b Igdm for every 9 E LI [a, bJ. •

6.3. THE SPACE OF FINITELY ADDITIVE SET FUNCTIONS

6.3

223

The Space of Finitely Additive Set Functions

In this section we describe the dual of LOO as a space of bounded, finitely additive set functions. Let A be an algebra of subsets of a set S. We denote by ba(A) the space of all bounded, finitely additive set functions on A. If we define iddition and scalar multiplication pointwise on A, ba(A) is a vector space, and if we define v <:: Jl to mean YeA) ::; /-LeA) for all A E A, then ba(A) becomes an ordered vector space; it follows from Lemma 3.12.1.1 that ba(A) is a vector lattice under the pointwise order. We define a norm on ba(A) by (6.1) Ilvll = Ivl(S), where Ivl is the variation of v [§2.2.1; that (1) defines a norm on ba(A) follows from Exercise 2.2.1.1J. From Proposition 2.2.1.7 (iv) it follows that (1) defines a lattice norIn so ba(A) is a normed lattice. We show that ba(A) is a Banach lattice. Theorelll 1 ba(A) is complete. Proof: Let {vd be a Cauchy sequence in ba(A). Let (> O. There exists n such that k, j :::. n implies (6.2) Then IVk(A) - Vj(A)1 < E for k, j :::. n and A E A [Proposition 2.2.1.7 (i)J so {vk(A)} converges, to say v(A). Fixing k :::. n in (2) and letting j - t 00 gives h(A) v(A)1 <:: E for A E A. By Proposition 2.2.1.7 (v) IIVk - vii::; (for k :::. n so v E ba(A) and Vk - t

v.

The Dual of LOO(/-L): Let /-L be a measure on a o--algebra, E, of subsets of S. Let ba(E : 1') be the subspace of ba(E) which consists of those v E ba(E) which are absolutely continuous with respect to /-L in the sense that fleE) 0 implies Ivl (E) O. Then ba(E : /-L) is a Banach space under the norm in (1) since ba(I:: : /-L) is a closed subspace of ba(E) [Exercise 1]. We show that each v E ba(I::: /-L) induces a continuous linear functional on LOO(p). We write IIfILX) for the fl·-essential sup of I and Ilfll" = sup{ I/( t)1 : t E S} for the sup-norm. If v E ba(E} and h : S - t R is bounded and I::-simple, the integral Is hdv is well-defined and (6,3) [Remark 3.2.2J. If h : S - t R is bounded and E-measurable, there exists a sequence of I::-simple functions {'f'h} such that 'f'k - t h uniformly on Sand I'f'kl ::; Ihl (3.1.1.3). From (3), Us 'f'kdv} is a Cauchy sequence so we may define Is hdv lim Is 'f'kdv; it is easily checked that the definition of Is hdv does not depend on the sequence {'f'd and (3) still holds [Exercise 2].

CHAPTER 6. FUNCTION SPACES

224

Lemma 2 Let v E ba(L: Jl) and h : S ~ R a bounded, L-measurable function such thatA {tES:h(t)#O} is Jl-null. ThenIshdv O. Proof: lIs hdvl $ Is Ihl d Ivl IA Ihl d Ivl + IS\A Ihl d Ivl $ Ilhll" Ivl (A) O. If v E ba(L : Jl) and h E LOO(Jl), we may define the integral of h with respect to v as follows: Pick f : S R bounded and L-measurable such that IIf - hll oo = 0 and IIhIL", = IIfll" [Exercise 3]. Now set Is hdv Is fdv; by Lemma 2 this definition does not depend on the choice of f and, moreover,

lis hdvl

$ Ilhll oo Ivl (S).

Hence, if v E ba(L : Jl), I,hen F,,(h) = Is hdv defines a continuous linear functional F" on LOO(Jl) with IIFyll $Ivl (S) Ilvll. It is actually the case that IIF" II = Ilvll. For suppose f > 0 is given. Choose a partition {E j : 1 $ j $ n} of S such that n

:L Iv(E)) I > II vii

f

j=1

[Proposition 2.2.1.7], and for each j let tj = signv(E)) and set h hE LOO(Jl), Ilhll"" $ 1, and n

:Llv(Ej )l2:: Ilvll-

f

j=l

so IlFvll 2:: Ilvll and equality holds. Thus, the map U : ba(L : Jl) ~ Loo(Jl)', U(v) = F", is a linear isometry which obviously preserves order. We show that U is onto so we have the Riesz Representation Theorem for LOO(Jl)'. Theorem 3 (Riesz Representation Theorem) U"'(Jl)' lattices}.

= ba(L : Jl) [as Banach

Proof: Let F E J/
£00

If L = peN), we write ba= ba(P(N)). If Jl is counting measure on peN), then LOO(Jl) and ba = ba(P(N) : Jl) so (f=y and ba are isometrically isomorphic as

Bana.ch lattices.

6.3. THE SPACE OF FINITELY ADDITIVE SET FUNCTIONS

225

Exercise 1. Show ba(2: : p,) is a closed subspace of ba(2:). Exercise 2. Show Is hdv is well-defined for h bounded and measurable and v E ba(2:) with lIs hdvl ~ IIhll ... Ivl (S). Exercise 3. (a) Show h E LOO(p,) if and only if :3 a bounded, 2:-measurable function f such that {t E S: h(t) i= f(t)} is p,-null. (b) Show that if hE L=(p,), t,hen Ilhll=

inf{llfllu:

and this inf is attained. [Hint: Show Ilfll" Ih(t)1 ~ IIhll oo } and show IlhCzll .. = IIhlloo']

f as in (a)} ~

Ilhlloo

for such f; set Z = {l E S :

Exercise 4. Let B( S, 'LJ be the space of all bounded, 2:-measurable functions. Show B(S, 2:) is a Banach space under the sup-norm and its dual is ba(2:). Exercise 5. Show Is hd(p, hE B(S,L,).

+ v)

Is hdp,

+ Is hdv

when

p"

v E ba (2:) and

226

6.4

CHAPTER 6. FUNCTION SPACES

The Space of Countably Additive Set Functions

Let 2:: be a u-algebra of subsets of S and let ca (2::) be the space of all finite signed measures on 2::. Since any finite signed measure on a u-algebra is bounded [2.2.1.5], ca (2::) is a linear subspace of ba (2::) and IIvll = Ivl (S) is a norm on ca (2::). We show ca (2::) is complete under this norm. Theorem 1 ca (2::) is a Banach space. Proof: We show that ca (2::) is a closed subspace of ba (2::) [§6.3]. Let {Jtd C ca (2::) be such that {Jtk} converges to Jt E ba (2::). Then

for each E E 2:: [2.2.1.7] so limJtk(E) from 2:: and set B

UB

j •

Let

f.

J.L(B) exists. Let {Bj

> 0 and choose

}

be pairwise disjoint

n such that lIJ.Ln -

j=l

IJt(B)

t

J.L(Bi)

+ I.

f

I~ IJ.L(E)

I=p+l

J.Ln(B)1

J.Ln(Bi)1 < 2£ +

+ It(J.Ln(Ei ) -

I. f

Jtll < L

Jt(E;)

Then

I (6.1 )

pn(Bi)l·

I=p+l

The last term on the right hand side of (1) can be made < £ for p large so Jt is count ably additive. ca (2::) is an ordered vector space under the pointwise order which it inherits from ba (2::), and we show that it is actually a vector lattice [Banach lattice]. Theorem 2 ca (2::) is a Banach lattice. Moreover, if p, v E ca (2::), then P 1\ veAl = inf {Jt(E)

+ v(A \B) : E E 2:, E

and P V veAl

= sup {Jt(B) + v(A\B) : E

E

C A}

2:, Be A}

forA E 2::.

Proof: By Lemma 3.12.1.1, it suffices to show that p 1\ v (Jt V v) is count ably additive. Let {Ai} C 2:: be pairwise disjoint and set A then

peE) + v(A\E)

UAi.

;=1

00

00

1=1

.=1

If E E 2::, E C A,

= 2: {Jt(Ai n E) + V(Ai\E)} ;::: 2: Jt 1\ V(Ai)

64. THE SPACE OF COUNTABLY ADDITIVE SET FUNCTIONS so

227

00

Jl A v(A) ::::

E Jl A v(A.) . • =1

Let

€

> O.

For each i there exists E. E

Jl(E.)

:L such that E.

C Ai and

+ V(Ai\E;) < Jl A v(A.) + €/2'.

Now {Ei} are pairwise disjoint and if E = so

UE"

then E C A and

,=1

U(Ai\Ei)

A\E

1=1 00

Jl

A

v(A) S; Jl(E)

Hence, Jl A v(A) =

+ v(A\E) = E{Jl(E;) + v(A,\Ei)} < EJl A v(A,) + €.

E

Jl A v(A i ). ,=1 The other statement is similar. There is an interesting order theoretic characterization of singularity for measures.

Proposition 3 Let Jl, v E ca (:L) be measures. Then Jl J.. v if and only if p A v Proof: =>: Let A, B E

a S; p A v(S) so Jl A v = O. <=: Jl A yeS) =

a implies

veEr) < 1/2'. Let An

0, S = AU B with Jl(A)

An B Jl

A

YeA)

+ Jl A v(B)

+ v(B)

that for each i there exists Ei E

n

= .UE i , A = n=l An = limE" 00

v(B)

O. Then

0

:L such that peE,) +

and B = AC. Then

00

Jl(A) S; Jl(An) S; EJl(E,) S; implies Jl(A)

S Jl(A)

=

= o.

E 1/2i = 1/2n-

1

O. Also

v(A~) = v eEl Ei) for i 2>: n implies v( A~) =

S; veEi')

< 1/2i

a so

v(B)

= v COl A~)

S;

~ v(A~)

O.

Hence, Jl J.. v. Exercise 1. Let A E ca (:L) be positive and let ca (:L : A) be the subspace of ca (:L) consisting of the elements which are absolutely continuous with respect to A. Show ca (:L : A) is isometrically isomorphic to Ll(A). Exercise 2. Show v E ba (:L) is purely finitely additive if and only if vJ..ca (:L).

228

CHAPTER 6. FUNGTION SPACES

6.5

The Space of Continuous Functions

Let S be a locally compact Hausdorff space and B B(S), the Borcl sets of S. Recall that Ge(S) is the space of all real-valued continuous functions on S which have compact support. We equip Ge(S) with the sup-norm, II . Ge(S) is not. generally complete [see Exercise 3J. Under the pointwise order, Ge(S) is a normed vector lattice. We begin by considering positive linear functionals on GeCS). If p is a Borel measure on B [so It is finite on compact subsets], then p induces a positive linear functional, FI" on (S) defined by < FI" <.p >= f" <.pdp. If F : GeCS) -> R is a positive linear functional on Ge(S) and F = f:' for some Borel measure ft, we say that JI is a representing measure for F. We first observe that regular representing measures arc unique. Proposition 1 Let ft, v be regular Borel measures on B such that for every <.p E Ge(S). Then p v.

Is <.pdp

=

Is <.pdv

Proof: Notation: If V is open and f E Ge(S), then f -< V means 0 ~ f ~ I and spt(f) c V. If K is compact and f E Ge(S), !..L-<J. mean;o~ f ~ 1 and f = 1 on

K. By regularity, it suffices to show that p(K) = v(K) for every compact K. Let e > O. Pick V open such that f{ c V and Jl(V) < Jl(J<) + Co There exists f E Ge ( S) such that K -< f -< V [3.5.5]. Since GK ~ f ~ Gv, p(V) < ,u(K)

+ I'.

so v(K) ~ p(K). By symmetry, v(K) J.t(K). Without the regularity assumption, Proposition 1 can fail; see Exercise ,1. We now show that every positive linear functional on Ge ( S) has a (unique) regular representing measure. For the proof we require a preliminary lemma.

Lemma 2 Let K

c

S be compact and Vi, ... , Vn open with K C

exist continuous functions fI,'"

(i)

O~li~l,i=l,

(ii) spt(Ji) (iii)

t

,=1

,In E Gc(S)

... ,n,

c \Ii,

J;(t) = 1 faT all t E K.

{f;} is called a partition of unity.

U\Ii.

1=1

such that

Then there

6.5. THE SPACE OF CONTINUOUS FUNCTIONS

229

Proof: Let x E K. Then x E V; for some i and there exists an open neighborhood Ux of x such that fl. is compact and x E Ux C U xcV;. Let XI, . .• ,X m E f{ be such that

KeD Ux ; and let Hi be the union of those Ux } such that UXj c Vi. By 3.5.5 1=1

for each i there exists 9. E Cc(S) such that Hi -< 9. -< V;. Put

Then fi -< V; and

(6.1 ) Since K C {) Hi, at least one 9i(X)

= 1 for each x

E

f{

so (iii) holds by (I) .

• :=1

To motivate the construction of a representing measure for a positive linear functional, we give the following result.

Proposition 3 If J.t is a re9ular Borel measure and V is open, then

Proof: Clearly JL(V) is greater than the sup above. Let r < /l(V). Choose K C V compact such that JL(K) > r. Choose f E Co(S) such that K -< f -< V [3.5.5]. Then Is fdJL ;::>: JL{K) > r so the equality follows. Theorem 4 (Riesz Representation Theorem) Let F be a positive linear functional on Cc(S). Then there exists a unique regular Borel measure JL representing F, i.e., F

F"..

Proof: For V open, define

JL(V)

= sup{F(f) : f

E Gc(S),

f -< V}

[ef. Proposition 3). Certainly, 0 S; JL(V) S; 00 and JL{U) S; JL(V) if U is open and U C V. Extend JL to all subsets of S by setting JL·(A) = inf{JL(V) : V open, V :) A}i note JL·(V) = JL(V) for all open V. We claim that JL. is an outer measure on S. Clearly 0 S; JL·(A) S; 00 for A C S, JL·(0) = 0 and JL. is increasing on subsets of S. It remains to show that JL. is count ably subadditive. Let {Ai} C S and A prove so assume

f

JL·(A i ) <

00.

00

00

i=l

;=1

U Ai. If E JL"(A.) =

00,

there is nothing to

Let (; > O. For each i choose V; open such that

1=1

Ai C V; and JL(V;) < JL·(Ai)+€/2 i • Set V

UVi,

i:1

If f -< V, then K

00

spt(f) C U Vi ;=1

CHAPTER 6. FUNCI'JON SPACES

230

so there exists n such that K C

UVi.

By Lemma 2 there exist fl,' .,in E Gc(S)

t=l

such that j, -<

Vi and

t j, = 1 on K. Then f S t j, so

i=l

i=l

n

n

F(J) S LF(J;} S

p(Vi)

00

00

1=1

t=]

s LJL(Vi) S LIL*(Ai) + (

1=1

and 'L(V) S L 'l*(A.)

+ {.

i=1

Hence,

,=1 Next, we claim that JL*(K) < 00 for each compact K C S Choose V open, with compact closure such that K C V. There exists 9 E Ge(S) with V -< 9 [3.5.5]. If f -< V, then f 9 so F(J) S F(g) and Il*(K) S p(V) S F(g) < 00. We now claim that p* is finitely additive over the family of compact sets. Let K I , Kz be compact with KI nK. = 0. Since p* is count ably subadditive, it suffices to show that I.L*(KI) + p*(Kz) S Il*(KI UK.). Since KI C K~, there exists an open VI with compact closure such that KI C Vi C VI C K 2. IIence,]{2 C (VIr· Set Vi = (VIr and note Vi n \.-2 = 0. Choose an open V :J KI U K2 with JL(V) S p'(I<1 U ]{z) + c. Then KI C V n Vi, K. C V n \.-2. Pick fl' h E Gc(S) such that fl -< V n VI, fz -< V n \.-2, p(V n Vi) < F(ft) + ( and p(V n Vi) < F(h) L Note ft + f2 -< V since Vi n Vz = 0. Then

s

p*(KI) + JL*(Kz) S JL(V n Vd + II(V n Vz) < F(Jd + F(Jz) + 2f = F(ft + fz) + 2( S p(V) + 2c < Il*(KI U K 2) + 3(

so 11'(Kt} + 'l*(K2) S p*(KI U Kz). We now claim that for each open V, 'ltV) = sup{p'(K) : V :J ]{ compact} lirHlcr regularity]. Let r < p(V). Choose f -< V with r < F(J). Let J{ spt(J). If Hi is open, K C Hi, then f -< Hi so r < F(J) S p(Hi) which implies 1.( V) ::::: p*( K) ::::: F(J) > r from which the desired equality follows. We next assert that B C M (,L*), the class of p* -measurable sets. First note that JL*( K) + 'L(V) = JL" (K U V) when K is compact, V is open and J{ n V = 0 [if J{ I C V is compact, then by the finite additivity of p* over compact sets and the countable subadditivity of JL*,

and the identity follows from the inner regularity of the preceding paragraph]. It suffices to show that if V is open, then V is p"-measurable. Let A C S with JL·(A) < 00 be a test set. First assume that A is open. Let K be compact with

6.5. THE SPACE OF CONTINUOUS FUNCTIONS

231

0 and An vc eWe A

K C AnV. Then the open set W A\K satisfies Knw so by the observation in the paragraph above,

The inner regularity of open sets implies that p:(A n V) +JL·(A \ V) ::; JL-(A) for every open A. Now let A c 5 be arbitrary. If W ::> A is open, then by the above

By the definition of JL-,

and V is JL--measurable. It follows that JL- restricted to B, which we denote by JL, is a measure on B. By the properties established above JL is a Borel measure and every open set is inner regular. Every Borel set is outer regular by the definition of JL- so JL is a regular Borel measure. Finally, we claim that F(f) = Is fdJL for f E Cc(S). Let f E Cc(S), E > 0 and fix V open with K = spt(f) C V and JL(V) < 00. Choose e> 0 such that If(t)1 < e for t E S. Partition the interval [-e,e] in the range of f by Yi = -e + i(2e/n), i 0,1, ... , n, where n is chosen such that 2e/n < E. Hence, Yi - Yi-l 2e/n < E. Let A = {t E K : Yi-l < f(t) ::; Yi} (i = 1, ... , n) be the partition of K induced by the {Yi}. Note the open set Wi = {t E V : Y'-l E < f(t) < Y' + c} contains Ai' By regularity for each i there exists open V; with Ai c V; C Wi and JL(V;\A i ) < E/n. n

Then K C U V;

.=1

and

c

V. By Lemma 2 there exists Yl, ... ,y", E Cc(S) such that Y; -< V;

Yi(t) = 1 for t E K. Note that fy; ::; (Yi F(f) -

t

Is fdp

i=l

<

F(y!;) -

+ E)Y; and

f

t

JYi. Thus,

i=l

Ik fdp I

t (Yi + c)F(gi) - t (Yi -

1=-1

i=1

.=1 '"

.=1

" + f)p(V;) - 2:(y, " < 2:(Yi

f)JL(A,) f)JL(A,)

n + f)(JL(V;) - JL(A,)) + 2E i=1 2: peA;) ~ t(c + E)f/n + 2EJL(K) = c[e+ f + 2p(K)] 1=1

2:(Yi

i=l

so F(f) ::;

Is fdp.

Replacing f by - f gives the reverse inequality.

Corollary 5 F is continuous if and only if p is finite. In this case,

Ilpll·

I!PII ::; peS)

232

CHAPTER 6. FUNCTION SPACES

Proof: "¢=: IF(J)I :s; Is If I dp, :s; IIfll oo p,(S). p,(S) sup{F(J): I -< S} :$IIFII since 11/1100 :s; 1 when f -< S. We denote by rca(S) the space of all finite, regular signed measures v on B with the norm Ilvll Ivl (S) [§6.4j recall a signed measure v is regular if Ivl is regular]. Each v E rca(S) induces a linear functional Fv on Ge(S) via integration,

'*:

(see Exer. 3.2.31), and since

IFv(J) I :s; £Ifl d Ivl :$ Ilfll oo IIvll , Fv is continuous with IlFvll :s; Ilvll. We say that v is a r'epresenting signed measur'c for Fv' Thus, we have a linear map U : v -+ Fv from rca(S) into Ge(S)'. We show that U is an isometry. Lemma 6 Let K 1 , ••. , Kn be pail'wise disjoint compact subsets of Sand ai, ... , an E R. Then thcl'e exists f E Gc(S) such that

(i) f(t)

a;jortEK;,

(ii) IIflloo IS

I

max{lad,···, lanl}·

Proof: There exist pairwise disjoint Vi, ... , Vn such that Ki C Vi [for n 2, this use induction). By 3.5.5 there exist Ii E Ge(S) such that J(i -< fi -< Vi, Set

ad.·

Theorem 7 Let v E rca(S). Then IIFIIII = IIvll. Proof: The inequality Ilf'vll :$ IIvll was observed above. Let t: > O. There exists a partition {AI," . ,An} of S by elements of B such that Iv(A.)1

>

IIvll-

t:

[2.2.1. 7). By regularity, choose compact sets K; C Ai such that n

IIvll

t:

<

L

n

Iv(Kdl :$

.=1

where we may assume v(K;)

f

= sign V(Ki) on

1= 0 for

all i. By Lemma 6 choose

= i=l UK i .

t 1 fdv t =

K,

Ivl (K;),

i==1

K. and IIfIL", :s; 1. Set K

i=1

L

i=1

Iv(Kdl

f

Then

>

Ilvll - (

E Ge(S) such that

THE SPACE OF CONTINUOUS FUNCTIONS

6.5.

233

and

Hence,

so

IlFvll

~ 11 11 11. We next show that U is a linear isometry onto C«S)'.

Theorem 8 U : rca(S)

-t

Cc(S)' is onto.

Proof: Let F E Cc(S)'. Then F = }; - F2 , where Fi are positive, continuous linear functionals on Ce(S) [Remark 5.7.20]. Let Ili be the regular Borel measure representing Fi [Theorem 4]. Since Fi is continuous, Ili is finite so Il III -1l2 E rca(S) is a representing signed measure for F. Thus, as Banach spaces rca(S) and Ce(S)' are the same. We show that they are also equal as Banach lattices, i.e., that U is a lattice isomorphism. Proposition 9 II E rca(S) is positive [i.e., a measure] if and only if Fv is positive. Proof: =;.: Clear. ..;=: By the first part of the proof of Theorem 4 the representing measure for Fv satisfies II(V) = sup{Fv(f) : f -< V}

°

°

for V open. Hence, II(V) ~ for every open V and II ~ by regularity. Thus, both U and U- I are positive linear maps and U is a lattice isomorphism by Theorem 5.7.9. Summarizing, we have Theorem 10 (Riesz Representation Theorem) U is an isometric lattice isomor-

phism from rca(S) onto Ce(S)' [rca(S)

= Cc(S)'].

Concerning reflexivity, we have Example 11 0[0,1] is not reflexive since 0[0,1] is separable but rca[O,I] is not separable [if t '" s, IIbt - b,lI = 2 where bt is the point mass measure at tj Corollary 5.6.1.13]. For the case when S is an interval I = [a, b] in R there is another description of the dual of O(I) using the Riemann-Stieltjes integral. If 9 is a function of bounded variation on I, then Fg(f) = fdg defines a linear functional on 0(1), and since

J:

IFg(f}1 ~

Fg

Ilfll oo Var(g : 1),

is continuous with IIFgl1 ~ Var(g : 1). It can be shown that IIFgl1 Var(g : 1) 80 the map \II : 9 - t Fg is a linear isometry from BV[a, b] into O(I)'. This isometry is not one-one, but if BV[a, b] is replaced by the space of normalized functions of bounded variation, N BV[a, b], consisting of the right continuous functions which vanish at a,

234

CHAPTER 6. FUNCTION SPACES

then W is a one-one, linear isometry from NBV[a,b] onto C(!)'. For a description of this version of the Riesz Representation Theorem, see [RN] or [TL]. For a discussion of the history of the Riesz Representation Theorem, see [Gr].

Exercise 1. Let xES. Define F : Ce(S) -> R by F(J) = J(x). Show F is a continuous, positive linear functional on Cc(S). What is its representing measure? Exercise 2. Let 9 : [a, b]

J: J

gdm. Show G E

-> R be continuous. Define G : CIa, b] CIa, by. What is G's representing measure?

->

R by G(J) =

Exercise 3. Let N have the discrete topology. Show Cc(N) = coo. Define F

Cc(N) by F(J)

f

j J(j).

OIl

Show F is positive, linear and find its representing

j~l

measure. Is F continuous?

Exercise 4. Let Rd be the real line with the discrete topology. Let S with the product topology. (a) Show that a subset of S is open if and only if its intersection with every vertical line is open in R. (b) Show the topology on S is locally compact. (c) Show J E Cc(S) if and only if J(.,y) E Ce(R) for every y and J(.,y) but finitely many y. (d) Define F on Cc(S) by: if J E Cc(S), let Xl,"" xn be those values there exists a y with J( x, y) i= 0 [part (c)] and set

F(j)

n

X

0 for all [or which

1: J(xi,y)dy = LR1: J(x,y)dy. xE

Show F is positive and linear. (e) Define

11

on B( S) by II( E)

L

m({y

(x,y) E E}). Show II

IS

a measure

xER

which represents F.

(f) Show

11

is not regular.

Exercise 5. Let C(X)(S) be the space of all continuous functions which vanish at 00 in the sense that for every € > 0 there exists a compact set K such that If( t)1 < c for t tI- K. Show cootS) is a Banach space under the sup-norm and describe its dual space.

235

6.6. HILBERT SPACE

6.6

Hilbert Space

In this chapter we set down the basic properties of Hilbert spaces. Hilbert spaces are special cases of Banach spaces and have many important and special properties not shared by general Banach spaces. Let X be a vector space over the field F of either real or complex numbers. If z E C, the complex conjugate of z is denoted by z and the modulus by Izl. Definition 1 An inner product (scalar product, dot product) on X is a function .: X x X --> F, (x,y) --> x· y, satisfying

(i) (x

+ y) . z

= x· z

+ y. z

Vx, y, z E X,

(ii) >.(x· y) = (>.x). y Vx, y EX,

>.

E F,

(iii) x· y = y. x Vx, Y E X, (iv) x· x 2: 0 Vx E X,

(v)

X·

x

=0

if and only if x

=

o.

A vector space X with an inner product defined on it is called all inner product space. It follows easily from the axioms that 0 . x = x ·0 = 0, x . (>.y) = :\(x . y) and X· (y + z) = x . y + X· z. We have the important inequality. Theorem 2 (Cauchy-Schwarz Inequality) If X is an inner product space, then

(6.1 ) Proof: If y = 0, the result is trivial so assume y

to

Ix .y / ~I ~ ..;x:x so we may assume y . y o~

=

1= O.

In this case, (1) is equivalent 1. Then

(x - (x· y)y). (x - (x· y)y) = X· x + Ix· Yl2 - (x· y)(y. x) - (x· y)(x. y) = x . x + Ix· Yl2 - (x . y)(x· y) - Ix· Yl2 = X • x -Ix· YI~ .

(6.2) Remark 3 Equality holds in (1) if x and yare linearly dependent. The converse also holds for if equality holds in (1) with y 1= 0, then (2) implies that x - (x· y)y = o. Note that axiom (v) was not used in the proof of (1). Proposition 4 If X is an inner product space, the map x norm on X.

-->

..;x:x = Ilxll

defines a

CHAPTER 6. FUNCTION SPACES

236

Proof: Only the triangle inequality needs to be checked. For x, y E X,

Ilx + yliZ

(x

+ y). (x + y) -:;

IIxliZ

+ IlyllZ + 211 xllllyll

(IIxil + IIYII)'

by Theorem 2. If X is an inner product space, we always assume that X is equipped with the norm induced by the inner product.

Proposition 5 The inner product is a continuous function from X x X Proof: If Xk

-+

x and

Yk

-+

F.

y, then

by the Cauchy-Schwarz Inequality. \Ve have the following important property of the norm in an inner product space.

Proposition 6 (Parallelogram Law) If X is an inner p1'Oduct space, then

Proof:

IIx+yllZ+lix =

yIl2=(x+y)'(x+y)+(x x .y

2lixl12 + 211yllZ + x . y + y . x -

y)·(x y . x.

y)

Remark 7 The parallelogram law characterizes inner product spaces among the class of NLS. That is, if X is a l\'LS whose norm satisfies the parallelogram law, then the norm of X is induced by an inner product. If X is real, the inner product is defined by 4x . y = lix + yliZ _ IIx yllZ, while if X is complex, the inner product is defined by

We leave the (tedious!) verification to the reader. See Exercise 7 for an example of a norm which is not induced by an inner product.

Definition 8 An inner product space which is complete under the induced no/'m is called a Hilbert space. Example 9 Rn or x· y

en is a Hilbert space under the usual inner product =

The norm induced by the inner product is the Euclidean norm.

6.6. HILBERT SPACE

237

Example 10 In the case of complex scalars, [2 is the space of complex-valued sc-

quences {Xi} such that

ex>

2

E Ix;! <

(x).

[2 under the inner product

i=l ex>

x·y

LXiYi,

X

{Xi},

Y

{Yi},

1=1

is a (the original!) Hilbert space. Example 11 Let (5, E, p) be a measure space. The space of real-valued functions, L2(p), is a Hilbert space under the inner product I· 9 = IsIgdp since this inner product induces the usual L2-norm. 'vVe also want to define the space L2(p) for complex-valued functions. For this we need to discuss integrating complex-valued functions.

Let I : 5 -+ C with u and v the real and imaginary parts of I, respectively. We say that I is measurable (integrable) if and only if both u and v are measurable (integrable). If I is p-integrable, we define the integral of I with respect to p to be

Is I dfl Is udfl + i Is vdfl· =

If Lb (fl) is the space of all C-valued, fl-integrable functions, it is easily checked that LI (fl) is a vector space and the integral is a linearfunctional onU (p). Since lui::; II I,

Ivl ::; III and III ::; lui + lvi, a measurable function integrable. Moreover, if I is integrable, then

I

is integrable if and only jf

III is

[let Is I dfl = lIs I dp I ei8 so

lIs Idfll == e- i8 Isldfl == I s e- i8 l dfl == Is R(e- i8 f)dfl +i Is I( e- i8 f)dfl Is R( e- i8 f)dfl ::; Is IR( e- i6 f) I dfl ::; Is III dfl, where Rz (Iz) denotes the real (imaginary) part of z E ej. For 1 ::; p < (X), let L~(fl) be the space of all C-valued measurable functions such that III E U(p) and set 1I/IIp = (Is IJI1' dp.)I/P. Then L~ is a complex Banach space; we usually write simply U(fl) for this space and indicate whether real or complex scalars are being used. For p 2, L2(p) over the scalar field of complex numbers is a Hilbert space under the inner product I . 9 Is Igdfl·

I

Example 12 C[a, b] with the inner product I· 9 space which is not a Hilbert space.

I: I(t)g(t)dt is an inner product

We establish an important geometric property of Hilbert space.

238

CHAPTER 6. FUNCTION SPACES

Theorem 13 Let K be non-void, closed, convex subset of a Hilbert space H. x E H, then there is a unique y E K such that IIx

zll: z E K}

yll = min{llx

If

dist(x,K).

Furthermore, y can be characterized by:

y E K, R(x - y) . (z - y) :$ 0 for all z E I<. Proof: Set d

dist(x,K)

~

(6.3)

O. If w, z E K, applying the parallelogram law to

(x - z)/2 and (x - w)/2 gives 2

d :$1I(w

+ z)/2 -

xll

2

= IIw -

since (w + z)/2 E K. If I/z xII = d and Ilw - xii Pick {yd ~ K such that Ilx

2

xll /2

+ liz -

2 2 x11 /2 -11(w - z)/211

(6.4)

d, then (4) implies that w z so uniqueness holds. Ykll - t d. Set w = Yk, Z = Yj in (fl) to obtain

Thus, {Yd is a Cauchy sequence in H and converges to some Y E K with IIx - YII If Y E K satisfies Ily - xII d, then for z E K and 0 < t < 1,

IIx-YII:$llx-tz-(1

t)YII

IIx-y-t(z

= d.

y)11

so

or

2R(x

y) . (z - y) :$ t liz _ YI12 .

Letting t - t 0 gives (3). On the other hand if y satisfies (3) and z E K, computing Ilx - zl12 = II (x gives IIx - Yl12 -llx - zll2 2R(x y). (z - Y) liz - yl12 :$ 0

y)

(z _ Y)112

so IIx yll = d. Let PK : H - t H be the "projection" map which sends x to y in Theorem 13. If H = R·, inequality (3) means that the vector from x to PKx makes an obtuse angle with the vector from z to PKx. Moreover, this map is uniformly continuous on H since Ilhu PKvll:$ lIu - vII· [from (3), R(u hu) . (hv hU):$ 0 and R(v-PKvHPKu-PKV) :$ 0 so adding gives R(U-V-(PKU-PKV))'(PKV-PKU) 'S: 0 so IIPKu PKvll2 'S: R(u v)· (hv - PKU) :$ lIu - vIIIIPKv PKul1 by the Cauchy-Schwarz Inequality.) If X is an inner product space, then two elements x, y E X are orthogonal, written xJ..y, if x . y = O. A subset E ~ X is said to be orthogonal if xJ..y y E x i y. If E ~ X is orthogonal and IIxll 1 vx E E, then E is said to be orthonormal.

6.6. HILBERT SPACE

239

Example 14 In £2 (over either R or C), {e k : kEN} is orthonormal, where ek is the sequence with a 1 in the kth coordinate and a in the other coordinates. Example 15 In L 2 [-1r,1rj (over C), {e int j..f2i; n = 0,

... } is orthonormal.

Proposition 16 (Pythagorean Theorem) If x.iy, then

Ilx + Yll2

IIxll 2+ Ily112.

yW

Proof: (x + y). (x + y) = Ilx + Yl12 = IIxl12 + IlyllZ = (x y). (x V). If X is an inner product space and M <;;: X, we set All. {x EX; x.iyVy E Af}; 11111. is called the orthogonal complement of AI. Note that 1..11. is a closed linear subspace of X. \Ve now use Theorem 13 to show that any closed linear subspace of a Hilbert space is complemented. Theorem 17 If lvlis a closed linear subspace of a Hilbert space II) then II M EB Ml.. Proof: For x E H let y PMx E M be as in Theorem 13. Then x - y E All. since by (3) R(x - y). w ~ a for every w E M. Ilence, x (x - y) + y with x - y E Ml. and y E M. Since lvf n All. {a}, we have II M EB Ml.. We can now establish the Riesz Representation Theorem for Hilbert space. Proposition 18 Let H be a Hilbert space and y E H. If fy : II (ly,x) = X· V, then fy E H' and IIfyll = Ilvll·

---->

F is defined by

Proof: fy is clearly linear and since l(Jy,x)1 = Ix· yl ~ Ilxllllyll, fy E H' with ~ lIyll· Since (fy,y) y. y = Ily112, IIfyll = ilyll· Thus, the map V ----> fy is an isometry from H into its dual space II'. We show that this map is onto.

Ilfyll

Theorem 19 (Riesz Representation Theorem) If II is a Hilbert space and f II') then 3 a 1J.nique y E H such that f fy. Proof: If f = 0, put y O. Suppose f # O. Set M = N(f), the kernel of f, so M is a proper closed subspace of Hand iU 1. # {a}. Choose z E M 1., Z # O. Then (/,z) # o. Set y ((f,z) j IIz112) z so Y E Ml., yolO, and (f,y) = 1(1, z)12 j IIzll2 = Y . y. For x E H, let

so x = Xl + Xz and (f, xI) a so Xl E M and Xl . Y O. Hence x • y = X2 • Y = (f,x) = (/y,x), and f fy. The map (> ; V ----> fy is an isometry from H onto H' which is additive but is only conjugate homogeneous in the sense that (>(ty) f(>(y). From this it follows that Hilbert spaces are always reflexive.

CHAPTER 6. FUNCTION SPACtJS

240

As noted earlier, L2 (p) is a Hilbert space so it follows from Theorem 19 that. the dual of L2 (p) can be identified with L2 (p) [in the case of real scalars]. This proof of the Riesz Representation Theorem for L2 (p) is independent of the proof given in Theorem 6.1.20 which depended on the use of the Radon-Nikodym Theorem. Indeed, an independent proof of the Radon-Nikodym Theorem can be based on Theorem 19; see [R2], Theorem 6.9. If X is an inner product space and E = {x" : a E A} is an orthonormal subscl of X, then Vx E X the scalars x(a) X· x a , a E A, are called the Four'ier coclJ!cienls of x with respect to E. We establish several important properties of the Fourier coefficients.

Proposition 20 Let X be an inner product space and {XI, . .. ,x n } an orthonormal set in X. Then for each X E X, n

(i) I:

.=1

(ii)

Ix' Xii

2

(X f: (x· Xi)Xi) l.xi Vi. 1=1

Proof: (i):

f:(x. Xi)X,) . (x ( X - '1::::::1

I:7=1 (x . Xi) Xi)

IIxl12 I:(x· ,

I:(x' x,)(X' X,)

+ I:I:(x· Xi)(X-;X;)Xi' Xi 'lIx112 - I: Ix . X;J2 . ,

(ii):

(x

Z:(X' Xi)Xi) . Xj

J

X· Xj - z:(x, Xi)(Xi' Xj)

= X·

x)

X·

x = O. J

We generalize the inequality in (i) to infinite orthonormal sets.

Proposition 21 Let E {xa: a E A} be an orthonormal set in an inner product space X. For each x E X the set Ex {a E A ; x . Xa of O} is at most c01mtabl£. Proof: For n EN, let

By Proposition 20 S" contains at most n

=

1 elements. Since E L · = US", the result n=l

follows.

Theorem 22 (Bessel's Inequality) Let E subset of an inner product space X. For each x

{x" ; a E A} be an orthono1'7nai

Xi (6.5)

aEA

aEA

6.6. HILBERT SPACE

241

Proof: If A is finite, this is Proposition 20. If A is infinite, we must assign a meaning to the series in (5). Let S {a E A : x . Xa. i= OJ. If S = 0, we set L Ix, xal 2 0, and if S is finite, we set aEA

nEA

aES

and (5) follows from Proposition 20. If S is infinite, S is countable by Proposition 21 so the elements {Xa : a E S} can be arranged in a sequence, say YJ,Y2, .... By Proposition 20, n

L

\In

Ix . yd ~ 2

IIxII 2

i=]

f

Ix . y;J2 is absolutely convergent and its sum is independent of the .=] ordering of the elements {xn : a E S}. Therefore, we may define

so the series

=

L Ix· xal 2 = L Ix· Yil

2

aEA

and

L Ix . xal

2

~

II xII 2

aEA

by Proposition 20. We will now show that equality holds in Bessel's Inequality for certain orthonormal sets in a Hilbert space. An orthonormal subset E of a Hilbert space}} is said to be complete (or a complete orthonormal set) if EJ c;;: H orthonormal and E J ,2 E implies that E = El (i.e., E is a maximal orthonormal set with respect to set inclusion). [See Exercise 1.] We give several criteria for an orthonormal set to be complete. First, we require a lemma.

Lemma 23 Let {Xl,.' ., X,,} be an orthonormal set in an inner product space X. n

(i) If X

L

CkXk,

then

Ck

X· Xk

=

x(k) and

k=l

(ii) For

{CI, ..•

,c,,}

F,

Ilx- EI

(Cl," .,c,,») at

Ck

=

Proof: (i): That

Ck

= x . Xk X • X

is immediate; =

LL k

ICkI2.

attains its mmlmum (as a function of

x(k), k = 1, ... ,no

X· Xk

IIxII 2 =

CkXk11

IIxI1 2 =

CkCjXk • Xj

242

CHAPTER 6. FUNCTION SPACES

(ii ):

o ~ Ilx - k~l CkXkf

and the expression on the right is clearly minimal at

Ck

x • xk.

Theorem 24 Let E = {x" : a E A} be an orthonormal set in a Hilbert space H. The following are equivalent:

(i) E is complete, (ii) xl..x" Va E A implies x

= 0,

(iii) span E is dense in H. (iv) If x E H, IIxll2 =

2

L:

Ix, x,,1 (equality in Bessel's Inequality),

"EA

(v) x

L: (x·

x")x,, VxE H,

aEA

(vi) if x, y E H,

X·

y

L: (x . xa)(y . x a)

(ParsC1Jal's Equality).

aEA

Proof: (i),*(ii): If (ii) is false, 3x # 0 such that xl..x a Va E A. Set z = xl IIxll so {z} U E is an orthonormal set which properly contains E so (i) docs not hold. (ii),*(iii): Let M be the closure of spanE. If lvJ # H, H = M ED M1. with lvJ1. # {OJ. If x # 0, x E lvJ1., then xl..x a Va E A so (ii) fails. (iii),*(iv): Let t> 0 and x E H. 3x ap ... ) Xan E E and Cl, .•. , Cn E F such that

By Lemma 23 (ii), (6.6) By Lemma 23 (i) and (6),

Bessel's Inequality gives the reverse inequality.

6.6. HILBERT SPACE

243

(iv)=>(v): As in Theorem 22 let S of S into a sequence Yt, Y2, . ... Then

{xa: Xa' x

:#

O} and arrange the elements

by (iv). Hence, 00

x

I::CX'Yk)Yk

.I:(X'Xa)Xa.

k=1

aEA

(v)=>(vi): By Proposition 5,

x .Y =

.I: .I:(x . aEAbEA

aEA

(vi)=>(i): If (i) fails, 3z E If with = 0 so (vi) fails.

L Iz· xal 2

IIzll

1 and z.Lx a Va E A. Then z· z = 1 while

aEA

Theorem 25 Let H be a Hilbert space with E and F complete orthonormal subsets. Then E and F have the same cardinality. Proof: Since orthonormal sets are linearly independent, we may assume that E and F are infinite. {J E F: f· e :# OJ. By Theorem 24 (ii), F = U 1". and by For e E E, let F. eEE

Proposition 21 each F. is at most countable. Hence, the cardinality of F is less than or equal to the cardinality of E. Symmetry gives the reverse inequality. The cardinality of a (any) complete orthonormal set is called the orthonormal dimension of the Hilbert space. Example 26 {e k : kEN} is a complete orthonormal subset of [2. More generally, let A be a non-empty set and let C a : A R be the characteristic function of {a} for a EA. Then {ca : a E A} is a complete orthonormal set in [2(A) [here [2(A) is L2(p) where p is the counting measure on A as in §6.1]. Theorem 27 (Riesz-Fischer) If H i.• a Hilbert "pace, then H i.• linearly isometric to [2(A) for .'lome A :# o. Proof: Let E {xa: a E A} be a complete orthonormal subset of H. By Theorem 24 the map U : H -+ f2(A) defined by Ux {x· Xa : a E A} is a linear isometry. Since U carries E onto the complete orthonormal set {c a ; a E A} in [2( A) [Example 26], U is onto peA). Note that the map U also preserves inner products by Theorem 24 so Hand peA) are isomorphic as inner product spaces. Example 28 {c'ntj.,jii; : n

£2[-11',11'].

= 0, ±1, ... }

E is a complete orthonormal subset of

CHAPTER 6. FUNCTION SPACES

244

This follows easily from condition (iii) of Theorem 24. By the Stone-Weierstrass Theorem, the span of E is dense in C[-1I",1I"] with respect to the sup-norm and C[-1T,1T] is dense in L 2[-1T,1I"]. Since convergence in the sup-norm implies converge in the L 2-norm, it follows that span E is dense in L2[ -1T, 11"]. If f E L2[-1T,1T] and

the series

I:

Ckeikt

is the classical Fourier series of

f [with respect to the complete

1.=-00

orthonormal set {e ik ! / V"ii : k E Z}] and the Ck are called the Fourier coefficients of f. If follows from Theorem 24 that this series always converges t.o f in the L2-norm. Note that the formula for the Fourier coefficient Ck is meaningful when f is integrable over [-1T, 1T]. One of the important problems in Fourier analysis is determining what integrable functions are such that their Fourier series converge to the function, at least a.e. duBois-Reymond gave an example of a continuous function whose Fourier series diverges at a single point. We will now show that the Uniform Boundedness Principle can be used to show the existence of such a function. Let f E Li [-1r, 11"]. The nth part.ial sum of the Fourier series for f is

= f:

eikt which appears in the integral above is called the Dirichlet kernel; we now compute a more useful form for Dn. We have (e it -1)Dn(t) ei(n+J)t e- int so e-it/2(eit _ l)Dn(t) = ei(n+J/2)t e-i (n+I/2)twhich implies

The function Dn(t)

k=-n

D,,(t)

sin(n + 1/2)t/ sin(t/2).

Thus,

sn(f)(t) = We define a linear functional

~ 211"

r

L"

f(s)Dn{t - s)ds.

}"'n on C[ -1r, 1r] by s,,(I)(O) = - 1

21r

i.e., F,,(f) is the

nth

1" f(s)Dn(s)ds, -7f

partial sum of the Fourier series for

f

evaluated at O. Since

Fniscont,inuousand IlFnll ~ 2~ IDn(s)lds. We claim that This equality follows easily from the following lemma.

IlFnll

Lemma 29 Let g : [a, b] -+ R be continuous. Define G : era, bJ -+ R by G(f) Then G is linear and continuous with IIGII Ig(t)1 dt.

J: f(t)g(t)dt.

J:

6.6. HILBERT SPACE Proof: Since

f: Ig(t)1 dt. Fix n

245

IGU)I ::; IIfiloo f: Ig(t)1 dt, G is

linear, continuous and

IIGII ::;

E N. Then

b 1a I9 I -- 1ba I9 11+"191 1+11191

J,ba ...-.W....rb 9 2fL. 1+,,191 + Ja I+nluj

::; f: ~ + G C;!lg)

::; b:a + IIGlllll;;luIIL", ::; ~!! + IIGII so

f: Ig(t)1 dt::; IIGII·

Next, we compute a lower estimate for f~1f IDn(s)1 ds. Since sin u ::; u for 0 ::; u ::; 7r,

{llFnl! : n

Thus,

E N} is unbounded. It follows from the UBP that there exists

f E C[ -7r, 7r] such that {Fn(J) : n} is unbounded, i.e., the Fourier series of f at 0 must diverge. Of course, the point 0 was chosen only for convenience, and the same result holds for any other point of [-7r, 1T] [the function f will depend upon the point chosen]. Banach and Steinhaus derived a method, called condensation of singularities, which can be used to construct a continuous function whose Fourier series diverges at any (arbitrary) countable subset of [-1T,1Tj. We first give their result. The proof uses the Baire Category Theorem [see A2]. Theorem 30 Let X be a Banach space, Y k a NLS and Tk E L(X, Y k ) for each kEN. Then B = {x EX: lim II Tkx II < oo} either coincides with X or is first catego,'y in X. Proof: Suppose B is second category in X. For each x E B, lim sup IItTnxl1 = k

Let

t

>

O. Then B C

UB

k,

{x EX: sup

where Bi;

k=l

n>!

o.

n;>1

IltTnxlis d.

Each Bk is

closed so some Bk contains a sphere, i.e., there exist xo- E X n B k , r > 0 such that Ilx xoll::; implies sup c. Thus, for /lzil 1', if = Xo + then

r

n;>!

IltTnxlls

s

and sup IITnzllS 2e.k. Hence, if x E X, sup n

n

IITnXl1

x

z,

S 2ek Ilxll/r so X = B.

CHAPTER 6. FUNCTJON SPACES

246

Corollary 31 (Condensation of Singularities) For each q E N let {Tpq}p be a sequence of operators in L(X, Yq), where X is a Banach space and Yq is a NLS. Suppose for each p there exists xI' E X with limll1~qxpll CXJ. Then B {x: lim II Tpqx II q

= CXJ Vp E N}

9

is second category in X.

= {x

Proof: For each p, Bp

EX: lim IJTpq x II q

Theorem 30 and the hypothesis. Thus, B

X\

<

CXJ}

is first category in X by

UBp is second category.

p=)

Let {til be a sequence of distinct points in [-7r, 7rJ. Let Snj(J) be the nth partial sum of the Fourier series of f E C[ -7r, 7r] evaluated at. tj. By the construction above, = 00. By Corollary 31 for each j there exists Ii E C[-7r,7r] such that. limlsni(Ii)1 n t.here exists f E C[-7r,7r) such that lim ISnj(J)1 00 for all j. That is, the Fourier n series of f diverges at each tj. lt can be shown that if the set {til is chosen to be dense in [-7r,7rj, then the set P = {t E [-7r,7r): lim JSn(f)(t)1 oo} is second category in [-7r,if] and, hence, uncountable [see [Y)II.4]. The classical Fourier series of an integrable function can behave very badly with respect to pointwise convergence. For example, Kolmogorov gave an example of an integrable function whose Fourier series diverges a.e. in [-if, if]. It was an open problem for many years whether the Fourier series of an L2 function must converge a.e. This was shown to be the case, only in 1966, by L. Carlcson. Hunt later extended Carleson's result to LP for 1 < p < 00. For a discussion of the results by Hunt, see

[As]. Whereas the classical Fourier series of integrable functions can display poor convergence behavior, the Cesaro averages of the partial sums of the Fourier series behave much better. For f E L1 [-if, if] the Cesaro averages of the Fourier series of fare defined to be series of

f.

0'"

(J)

=

f

Sk

(f), where Sn (J) is the nth partial sum of the Fourier

k=O

From the discussion above, 0'"

The function Kn (t) =

(f) (t)

f

1

-~

j"

1

~-~

27r -" n

n

+ 1 k=O

Ddl

s) f (s) ds.

Dk (I) is called the Pejer kernel. Thus, we have

k=O

O'n(l)(t) = 1

27r

j1f Kn(t-s)f(s)ds. -1f

If we assume that all functions in Ll [-7r,if] are extended periodically (with period 2if) to R and if we replace integration over R by integration over [-if, if], we may interpret the formula above as a convolution integral,

1

O'n(f) (t) = 2if

*f

(t)

6.6. HILBERT SPACE

247

as in §3.11. We will now show that the sequence {2~Kn} satisfies the properties of an approximate identity as defined in 3.11.2. For thIs it is convenient to derive another formula for the Fejer kernel. Substituting (e it - 1) Dn (t) ei(n+l)t e- int into the definition of Kn gives

(n

+ 1) Kn (t) (e it

1) (e- it - 1) = (e- it - 1) Lk=O (ei(n+l)t = 2 ei(n+l)t _ e-i(n+l)t

e- int )

so that Kn (t)

1 n+1

Kn (t) ~

(n+l)(;-co06)'

cos(n+1)t 1 cos t

(6.7)

From (7), we obtain

Lemma 32 (i) Kn :?: 0,

(ii)

J~1f Kn = 27r J

(iii) For 0 < 5 ~

It I ~ 7r,

From (i), (ii) and (iii), it follows that {trKn} satisfies the conditions of an approximate identity given in 3.11.2. [Note that the Dirichlet kernel fails property (i).] From the analogue of Theorem 6.1.24, it follows that if I E £P [-:>r,:>rJ, then {ern converges in LP- norm to I. Also, from the analogue of Exercise 3.11.10, it follows that if I is a continuous function on R of period 27r, the Cesaro averages (I)} converge to I uniformly on [-7r, :>r)i this result is known as Fejer's Theorem. 1: ., also the case that if I E Ll 7r], then {un (I)} converges a.e. to I; see [HS] V.li:l.29. For a discussion of the historical development of Fourier series by Zygrnund, see [As). Zygmund's book ([Zy]) contains a detailed discussion of Fourier series.

(In

Exercise 1. Show that any orthonormal subset of a Hilbert space is conta.ined in a complete orthonormal set. Exercise 2. Show a Hilbert space is separable if and only if its orthonormal dimension is countable. Exercise 3. If D is a dense subset of an inner product space and xl.D, show x

= o.

Exercise 4. If E is a linear subspace of a Hilbert space H, Y is a. B-spr.ce and T : E -> Y is a continuous linear operator, show T has a continuous linear extension T: H -> Y.

248

CHAPTER 6. FUNCTION SPACES

Exercise 5 (Gram-Schmidt). Let

XI,""

Xn

= Xl,

be linearly independent. Set. Yl

k-l Yk

=

Xk -

I)Xk' Yi)yJlIIYjI12 J=1

for k> 1 and

Zk

=

Yk/IIYkll.

Show {z".) is orthonormal and span{xd

span{zd.

Exercise 6 (Riemann-Lebesgue Lemma). Show that if! E L1 [-11"", 1I""J, thell

~ f" !(l)e-·ktdt

Ck:= V

as

Ikl -->

00.

211"" L~

-->

0

[Hint: First prove this for characteristic functions of intervals.J

Exercise 7. Show that the sup-norm on C [0, IJ docs not satisfy the parallelogram law. Exercise 8. Let M be a closed subspace of a Hilbert space II. (i) Show Ml.l.

M.

(ii) Show III M is a Hilbert space. (iii) If x E If, x m + m' where m E IVl, m' E l11.L as in Theorem 17, show I):]" = defines a continuolls linear operator on 11 sllch that p2 fJ.

In

Exercise 9. Show a linear isometry onto an inlier product. space preserves inner products. Exercise 10. Let lvI, 1'1 be closed subspaccs of II with M 1- N. Show At closed.

N is

Exercise 11. Show a linear subspace M of a Hilbert spaef' Is dellse if and only if M.L {O}. Give an example of a proper closed subspace lVi of all inner product space such that M.L {O}. Exercise 12. let E

{e. : a E A} be an orthonormal family in a Hilbert space ll.

Show the closed linear subspace generated by E is

{I:

aEA

t.e a

:

I: Ilal2 < oo}.

aEA

Exercise 13. Show TheAJrem 19 is false for inner product spaces. Exercise 14 (Hellinger-Toeplitz)Let T : II --> II be linear and satisfy Tx· y for all x, y. Show T is continuous. Hint: Use UBP or CGT.

= X· Ty

249

6.6. HILBERT SPACE

6.6.1

The Fourier Transform

In this section we define and study some of the basic properties of the Fourier transform. Although the Fourier transform is defined for functions on Rn, we restrict our attention to functions on R. In dealing with the Fourier transform, there is always a factor of 27T involved. We will take care of this by using the normalized measure jt m/...;z:i. We denote by IZ the space U(jt), where we are considering complexvalued functions. All statements about measurability refer to the Lebesgue measure, and any integral over R is denoted by f f dp = fR f djt. In this section, we agree that. the convolution of two measurable functions f and 9 is defined by

f

* g(x) =

Jf(x - y)g(y)dp(y)

[see §3.11]. If fELt, we define the Fourier transform of f, j, by ](t) f e- itx f(x)djt(x). it Note that since e- ", is bounded and is continuous as a function of x, j is defined for all t E R. We list some of the basic properties of the Fourier transform.

Theorem 1 Let

f,

9 E Ll

(i) Ifh(x)=eiaxf(x), thenh(t)=j(t (ii) If hex)

= f(x

a), then h(f)

= e-iatj(t). (i(t»-.

(iii) Ifh(x)

l(-x), then h(tl

(iv) If hex)

f(ax), a > 0, then h(f)

(v) (f

* g)"

(vi) If hex)

a).

= f(t/a)/a.

j9.

= -ixf(x)

and hE V, then j is differentiable and

(vii) If f' ELI and lim f(x) Ixl-oo

(1)' (t)

h(t).

0, then (fl)"(t) = itj(t).

(viii) (Riemann-Lebesgue Lemma) j is continuous, bounded lim Jet) = 0.

[iinioo ::; Ilflll]

and

Itl-oo

(ix) f f9

= f jg.

Proof: (i)-(iv) are easily checked. (v) follows from Fubini's Theorem [recall 3.11.1] SInce

(f

* g)"(t)

f e- itx f f(x - y)g(y)dp(y)dp(x) f g(y)e- itll f e-it(x-y) f(x - y)djt(x)dp(y) f g(y)e-itlldjt(y) f e- itx f(x)djt(x) = ](t)g(t).

250

CHAPTER 6. FUNCTlON SPACES

(vi) follows from 3.4.3. (vii) follows by integration by parts. (viii): is bounded with l!(t)1 :S IIIII1 for every t. Since e"i

I

!(t)

=-

Jl(x)e-it

(x+1f

j

t)dj.t(x) = -

JI(x -7r/t)e-

1, i1x

dj.t(x)

so

as t

-+ 00

by 3.11.10. Finally, for t, s E R, we have

l1(t + s) -1(t)1 :S and since limsin(.sx/2) 8->0

J

11 d/l(x)

I/(x)1

= 0 for

=

J

I/(x)112sin(sx/2)1 dx,

each x, the DCT implies that the last term is small

I

for s small independent of t. Hence, is uniformly continuous. (ix) follows from Fubini's Theorem. Conditions (vii) shows that the Fourier transform converts differentiation into multiplication by it, and this property makes the Fourier transform useful in the study of differential equations. Conditions (vi) and (vii) also point out another important property of the Fourier transform: growth properties of I arc reflected in smoothness properties of and vice-versa. From (viii), the Fourier transform is a continuous linear transform from I} iuto the space, Co(R) , of bounded, continuous fnnctions on R which vanish at 00 when Co(R) is equipped with the sup-norm. The Fourier transform is not onto Co(R) [Exercise 7]; there is not an intrinsic characterization of the functions which are Fourier transforms of U functions. {f : R -+ R : I is infinitely differentiable and for each j, k, sup{lxjll/(k)(x)l. Let S J.; E R} < oo}. S is called the space of rapidly deC1'easing lunctions or the Schwar·tz space after Laurent Schwartz. By Leibniz' rule for the differentiation of products, it is easy to see that an infmitely differentiable function I belongs t.o S if and only if (xl/(x))(k) is bounded for each j, k. Thp function I(x) furnishes an example of a function in S [see Example 3 below]; indeed, C;:'(R) C S so S is dense in LI (3.11.12).

I

Corollary 2 SA C S. Proof: If I E S, for each j, k the function x -+ xi I{k)(x) is integrable and vanishes at 00 so is infinitely differentiable by (vi), and by (vii),

I

IE S.

J

so ((it)kl(t)t is bounded by (viii). Hence, by the observation above, It will follow from the Fourier Inversion Theorem below that SA = S.

Example 3 If I(x)

= e- x2j2 ,

then

1=1.

6.6. HILBERT SPACE

251

From (vi),

(1)' (t) == and integration by parts gives

J_ixe- i:;;te-x /2dp,(x) 2

(1)' (t) == -t I( t).

Hence, the derivative of the quotient

c, a constant. Putting t == 0 gives c 1 from 3.4.4 in the measure]. [because of the normalization factor We now consider inverting the Fourier transform. For this we define the inverse (x) == Fourier transform of a function f E V to be (t) = f eit:r f (x )dp,( x). Since !( -x), the inverse Fourier transform shares most of the properties of the Fourier transform given in Theorem 1. We do not bother to record these properties. Let g(x) == e- r2 / 2 and ga(x) == e-(xa)2/2. By Example 3 and Theorem 1 (iv),

!(t)/ f(t) is 0 so I(t)/ f(t)

-Jr;

r

r

g,,(t) == (l/a)gl/a(t). Theorem 4 (Fourier Inversion Theorem) Let f,

1 ELI.

Then

(1) v == f

a.e.

Proof: By Theorem 1 (ix) and (i),

f !(t)eixtg" (t)dp,(t)

f f(t) (ei:rtg" (t»" dp,(t) == f f(t)g,,(t - x)dp,(t) =

Since limg" (x) <1-+0

= 1 and

f

f(t)~9I/,,(t

(6.1)

x)dp,(t)

f

* !gl/a(X).

the integrand on the left hand side of (1) is bounded by

Ill,

the DCT implies that the left hand side of (1) goes to v (x) as a --+ O. If k = ~, kEN, ~gl/a is the approximate identity in Example 3.11.5 so the right hand side of (1) converges in V-norm to f [3.11.11]. Since any sequence which converges in

(1)

L1 -norm has a subsequence which converges pointwise a.e. [3.7.1, 3.6.3],

(1) v

f

a.e.

1

Remark 5 Thus, if f, E Ll, then f can be made into a continuous function which v = f everywhere [Theorem 1 vanishes at 00 by modification on a null set and (viii)].

(1)

From the computation in Corollary 2, we have

Corollary 6 S" = S. Concerning uniqueness for the Fourier transform, we have

Corollary 7 If fELl and

1

0, then f

0 a.e.

We now consider extending the Fourier transform to L2j the resulting extension is sometimes called the Plancherel transform.

1

Lemma 8 Let X == {J E Ll : ELI}. If f E X, then 1 EX C L2 and the Fourier transform restricted to X preserves inner products on X.

CHAPTER 6. FUNCTION SPACES

252

Proof: By Theorem 4, X" X and if 1 E X, 1 is bounded so to L2 [Exercise 6.2.9J. If 1, 9 E X, set h = (ffr. Then

1 and

1 belong

almost everywhere. Hence,

so the Fourier transform on X preserves inner products. It follows from Lemma 8 that if 1 EX, then 111112 = 111112 so the Fourier transform is a linear isometry of X onto X. Since SeX, X is dense in L2 so the Fourier transform can be extended to a linear isometry from L2 onto L2 which preserves inner products. We need to show that this extension agrees with the Fourier transform on LI n P. Let 1 E Ll n L2 and let hk(t) = kgk(t) be the approximate identity used in the Fourier Inversion Theorem above. Then 1 * hi< -+ 1 in L I [3.11.11] so (J * hk )" -+ 1 uniformly on R [Exercise 1]. By Theorem 6.1.24, 1 * hi< -+ 1 in L2-norm so 1 is equal to the L2-extension of the Fourier transform of 1. We continue to denote the extension of the Fourier transform of a function 1 E L2 by Let 1 E L2 and set!k Cr-k,k)I. Then Ik 1 in L2 so - t 1 in P. Since Ik E LI n L2, its Fourier transform is given by

f.

h

and the Fourier transform of 1 is the L2-limit of the sequence {lk}' [A similar formula holds for the inverse Fourier transform.] ;\ote that 1 is only determined up to a.e. since it is an L2-limit whereas the Fourier transform of an Ll-function is unambiguously defmed everywhere.

Exercise 1. If

II.:

-+

1 in

LI, show

lk -+ 1

uniformly on R.

Exercise 2. If 1 E L2 and 1 E LI, show

for almost all x E R.

Exercise 3. If 1, 9 E L2, show (lff) v = 1 * g. [;\ote and 1 * 9 makes sense by Exercise 6.2.6.]

lff E £I

by Holder's Inequality

6.6. HILBERT SPACE Exercise 4. Let 'Pa

= CI_a,a]'

Show ;f>a(t)

= 2si~"t, II'Pall; = 2a, and f

~a. Exercise 5. If f(x)

= e- 1rl , find

Exercise 6. If f E £1 and f Exercise 7. Let

*f

(Biny"Y

r

dJ1(Y)

v

f. Show (f(xlk))V defines an approximate identity. = f[OJ a.e., show f = 0 a.e.

f E £1 be odd. Use the fact that

{It

r

:

lal < I,BI} is

bounded

and Fubini's Theorem to show that {f; f(t)ltdt : b> I} is bounded. Use this to show that g(t) = lien It I for It I :2: e and g(t) = 0 for Iti < e is not the Fourier transform of an Ll function.

A. L

FUNCTIONS OF' BOUNDED VAHIATIONS'

25.5

AI: Functions of Bounded Variations For the reader who may not be familiar with the basic properties of functions of bounded variation, we record them in this appendix. Let f : [a, bj -. R. If ir {a Xo < Xl < ... < In = b} is a of fa, b], the variation of f over ii is n-I

L

: iT)

If(.l:iH) - f(xi)I,

i=O

and the variation of f over [a, bj is

Var(J : [a, b])

= sup c'ar(J : )f),

where t.he supremum is taken over all possible pitftitiolls,

b]) <

of [a, bj. If

ir,

<Xl,

f is sajd to have bounded variation; the class of all such functions is denoted by llV[a, bl. The variation measures the amount the function oscillates in As the example below even a continuous function can fa.il to belollg to BY[a,bj. Example 1 Let f(t) Isin(1/I) forO < t::; I and frO) = o. Set Xn 1/(n+l/2)ir. Thell f(xn) = l/(n + 1/2)ir if n is eveIl, and f(xn) = -l/(n + 1/2)11" if n is odd. If 11"" is the partition {O <.rn < Xn-l < ... < :rl < I}, then ,,-1

') n-I

f(x,-Ill ~ ::

L

l/(i + 1)

Tt t=l

soVar(J:[O,l])

<Xl.

Proposition 2 If f E BV[a, bJ, then f is bounded on [a, bl· Proof: Let

X

E

b). Then

f(a:11 + If(b)

f(xll::; Var(J: [a,b])

so

2lf(:rJI

If(a)1

+ If(bJI + Var(j

: [a,

blJ

and f is bounded. We consider properties of Var(j : l) as a fUIlction of the interval T. we have consequence of the triangle

Lemma 3 Let f : [a, bj -. R. If : ii) ::; var(j . iiI).

ii

and

iii

are partitions of [a, b] with iT

as a

irl,

then

256

APPENDIX A.

Proposition 4 Let

f : [a, b]

-->

R and a < c < b. Then

+

VaTU: [a, b]) = VarU: [a,

b]).

Proof: Let ii be a partition of [a, b] and iii the partition obtained by adding the point c to 71. Let 711 and "2 be the partitioIls of c] and [c. b], respectively, iIlduced by 71 1• Then by Lemma :3,

varU:

71) :::::

varU : ii') = varU:

7Id + vaTU: 712)

:::::

Var'U : [a, cll

+ VaTU:

[c, b])

so

VarU : [a, b]) ::::: VarU . [a. cD

c] aud fe, b], r('spectivcly, then"

If "I and "2 are partitions of partition of [a, b] so varU: ,,)

=

vaTU:

+ VarU : [c, b]).

7[1)

+

"] U

7[2

is a.

bj).

Val'U: [a, cD + Var(f : [c. we consider how the variation, V nrU

Proposition 5 Let

(i) f

+9E

(ii) fort

: /), dep()nds

Oil

the fundion

f.

f. 9 E BFta, Ii]. Then

BV[a, b] with l/arU + 9 : [a. It]) ::::: VarU :

R,tfE T3V[a,b] withVar(tf:[a,bJl

b])

ItiVar(f:

Var(,q: [a, Ii]),

b]).

Proof: (i) follows frofll the triallgle illequality ilnd (ii) is clear. Thus, BV[a, b] is a vector space uuder the usual pointwis(' addition and scalar multiplication of functions. Let f E BI/[a, Ii]. We define thp tolal l1ar;otion of f by Vf(t) VaI'U: [a. tll if a
f arc

incnrt.sillg on [a, b].

Proof: Vf is increasing by Pl'Oposition .1. Let a .1' < y ::::: band g = 1'1 - f· Then

g(y)

=

llf(x)

+ Val'(f:

[:r,y])

f{y)

implies

"2:0

A.l. FUNCTIONS OF BOUNDED VAHIA'TIONS

257

Proposition 7 Iff E BV[a, b] is (right, left) continuous at x, then VI is (right; left) continuous at x. Proof: Let t> O. Suppose f is continllous at. ;T < b. There is a partition 7r of [x, b] such that varU : Var(f : t. Since f is right continuous at ;c, we may add a point Xl to 7r to obtain a partition

- f(xdl

of [.T, b] such that

f.

Then n-l

(+ var(f :

+ f(xdl +

so

F a/'(f : [x, bJ) <

: 7rt)

+ f < 2t + VaT(f:

[Xll

b]).

2(, and since Ff 1,

Thus, 0 S FJCrJ)

continUOl1s. The stat.ement about left is similar. Since f = Vf-(Vf - f), Propositions 6 and 7 along with Exercise I and Proposition 5 give the following characterization of functions of bounded variation.

Theorem 8 Let f : [a, b] -+ R. Then f E HI/[a, b] if and only if f = g - hi where g, Ii If f is (right, hft) continuous, then 9 and h ran be chosen to be (l'ight, left) continuous.

r.

Exercise 1. If f

i

on [a,b], show Fm'(,f: [a,b])

feb) - f(a).

Exercise 2. Give a necessary and sufficient condition for

VarU'

b])

f

to satisfy

O.

Exercise 3. Let j, 9 E BF(a, Show fg and If I belong to BF[a,b], and, hence, f 1\ g, f V g E EF[a, bl [formula 1.2 of . That HF[a, b] is a vector lattice and also an algebra; in the terminology of Example 5.7.7 HF[a, b] is a function space. y) :

Exercise 4. Let rp

7r

is a partition of

b] -, R2 be continuous. If

to

11

...

< (,

b}

b], set

L( rp, 7r)

Then L(rp) = sup L(rp, 70), where the supremum is taken over all part.itions 70, is the arclength of the path rp. Show L(rp) < CX) if a.nd only if both x, y E BV[a, b].

258

APPENDIX A.

A2: The Baire Category Theorem The Baire Category Theorem has been used several times in the text. For the convenience of the reader unfamiliar with this result, we give a statement and proof in this appendix. References to further applications of the Baire Category Theorem are given at the end of the Let (8, d) be a metric space. 1\ subset E c S is nowhert dense if the interior of F: is empty. For example, t.he Cantor set is nowhere dense in [0,1] [Exercise A subset E c S is in S if E is a countable union of nowhere dense sets. in R. A subset E c S is second category in 5 if E For example, Q is first is not first category in S. l3aires' Theorem asserts that every complete metric space is second category in itself. Proposition 1 Let {Pi-} be a sequence oj ci08td 8ds coniained in the complete melric space (5, d) such that 1'1,

and d k = diameter PI,

-7

O.

Then

n

Fk z.~ a

k=1

singleton.

-s: Ih for.i

Proof: It suffices to show so {J.J is Cauchy and, tiIen>fore, convergent to some

J'

E

.

S. Clearly;r E

k

nF".

k=1

Theorem 2 (Baire Category Theorem) 11 complete mcinc space is second category in itsc(f Proof: Let 11k be nowhere dense in S for every kEN and set E

UAk.

"=1

We show ~. Let Xo E S. Since Al is nowhere dense, there is a closed ball B1 of radius less than I inside the dosed ballBo = {x : d(x, To) I} such that Bl n Al = 0 1). Since A2 is llowheff' dense, there is a closed baH B2 of radius less than 1 inside Bl such that. B2 n A2 = 0. Continuing this construct.ion such that gives a decreasing sequence of closed balls {Bd of radius less t.han I

Bk n Ak

0 for all k·.

Proposit.ion 1,

nB"

=

{x}. Clearly x E

k=l it corollary of the l3aire We applications.

Corollary 3 ',el

el) be a

Theorem which is particularly useful in

metric space.

~f

S

lh,

then some

mllst have a non-void interior.

Despit.e its esoteric appearance the Haire Category Theorem has a number of applications to various areas of a,nalysis. For example, Banach used the t.heorem to show the existence of a continuous, nowhere differentiable funct.ion. For this alllI other interesting see [DeS] and [Boa,].

259

A.2. THE BAIRE CATEGORY THEOREM

Exercise 1. Show that E is nowhere dense if ilnd only if every sphere S contains il 8' such that 8' n E = 0. Exercise 2. Show thf' Cantor set is nowhere df'nse in [0,1]. Exercise 3. Let.!k : R ~ R be continuolls, nonnegative and such that

f

k=]

converges for every t E R. Show that there is an interval in R where the convergence is uniform. Exercise 4. Show that R2 is not a countable union of lines. Exercise 5. Show that if S is a complete metric space without. isolat.ed points, then S is 1lIlcountable.

260

APPENDIX A.

A3: The Arzela-AscoIi Theorem Let S be a compact Hausdorff space and assume has the sup~llorlU. vVe give a characterization of the compact subsets of C(S) due t.o Arzela and A,iCOIi. Of course, any compact subset of is closed and bounded, but by Theol'(,1Il .5.1.18 the converse cannot hold ill general. We firsl give an additional necessary condition that a compact subset of must satisfy.

Definition 1 A s'UiJsci 1{ 15 al s E S if for eve"y ( 0 then exists a nci.l/hborhood V of$ ,'judi lhallf(s) - fO)1 e every t E V and f r..... J{ is (on S) zf J{ ('quirontin1J01l8 al (VC1'Y point of S. Proposition 2 If]{ C CIS) is compad. thel1 I{ is Proof: Let ( > 0 and s where

IN,,)

f

E

If(l) - f(s)1 so

S. 'I'here exist ft .... ,fk E g such that K

,e) = {g : Ilg fll fi(t)1 < ( for t E V and i

such that

eqllicolliinuous.

Pick an open neighborhood V of oS such that l, ... ,k. Suppose I E V and f E I\'. Choose i

Then

If(t) - h(lll

+ If,(I)

f,(8)1

+ 1/,(8)

f(s)1 < 3(

is eqllicolltinuous. a.uy compact subset of C( S) is closed. bounded and cquicontiuuons. vVe cousider the converse of this statement. j(

Lemma 3 Let {fd be a sfquena of real-valued fUTlctions drjincd on n conntable 8ft E = {Tk : kEN} which is bounded OTl E. Then there is a subseqlLence l!7k} of Uk} such that L'lk} converges pointwise on E. Proof (Diagonalization Procedure): 'fhe sequence {fk(Td} is bounded in Rand, therefore, has a convergent subsequence {fl,k(Tl)}~l [the reason for this slightly Ull~ orthodox notation will become The sequencc }~=l is bounded and has a convergent snbseqlJ('nce . Xote that )} ~l also converges. Proceeding by induction produces a sequence {8d as follows:

81

f1,l

fl,2

hI h2 Note that (i) is a subsequence of and Now let {gd be the diagonal sequence {h.d.

{j~.dTi)

con verges for 1

:S i :S j.

Theorem 4 (Arzela-Ascoli) Let j{ C CIS). Then i{ is compact if and only if K is closed, bounded and equiconlinuo'lJ.s.

A.3. THE ARZELA-/18COLl THEORElvl

261

Proof: =;,: Proposition 2. ft suffices to show that any sequence {fA·} c K has a subsequence which eqllicontinuity, for each k there exists a finite set converges uniformly on S. Fi.: C S and open neighborhoods {lit : t E Fk } Sitch t,hat S = 11;: t E Fi.:} and If(s) - f(t)1 11k when E \.~ and f K. Set E

UFk. By Lemma a ther(' is a subsequence {gk} of {fA-} such that

{9k}

k=1

converges pointwise on E. \Ve claim that {gk} is a Cauchy sequence in Let c> 0 and choose k snch that 11k t. There exists IV such that Igi(t) gAt) I < 11k [or all t E Fk and i, j ::: N. If 8 E S, thert> exists t E such that Ii E Y~ so if i, J N,

lIence, {gk} is a Cauchy sequence in and the proof is complete. If f{ c C(S) is bounded and equicontinuous, then Ii is ('quimntinnou8 [Exercise 1) so is compact, by Theorem 1. Hence, every sequence in !,,' has subsequence which converges uniformly on S. The Arzela-Ascoli Theorem has ma.ny applica.tions in differentia'! equations, integral and cakulns of variations. See, for example, Exercise 1. IrK

is equi.;olltinllous, show K is eqllicontinnous.

Exercise 2. Give a specific example of a subset of but not compact.

e[o, 1]

which is closed, hounded

Exercise 3. Let J{ C b] be such that each f E l( has a continuolls derivative ilnd [(' {I' : f E I<} is bounded in bJ. Show J( is equicontinuous.

era,

Exercise 4. Let J{ c C[a,bj be bounded. Let. F(t) = is equicontinuous.

J: f(8)d5.

Show {F: f E [(}

Exercise 5. Let J( C be equicontinuous and pointwise bounded on S'. Shaw K is uniformly bounded on S. Exercise 6. Let {fd C be equicontinuous. If {f,J converges pointwise on a dense subset of S, show {fd converges pointwise OIl 8. Is compactness important?

262

APPENDIX A.

A4: The Stone-Weierstrass Theorem In this appendix we prove Stone's far reaching generalization of the Weierstrass Approximation Theorem. Let S be a compact Hausdorff space and C(S) the space of all real-valued continuous functions on S equipped with the sup-norm [complex-valued continuous functions are considered at the end of this section]. Stone's Theorem gives algebraic conditions, modeled on the polynomials on the line, which insure that a subset of C(S) is dense in C(S). A subset A <;;; C(S) is called an algebra if (i) f, 9 E A imply f

+ 9 and

fg belong to A.

(ii) f E A implies tf E A Vt E R. Example 1 C(S) is an algebra. The set of polynomials, 'P, is an algebra in C[a, b]. The set of even polynomials, [;, is an aJgebra in C[a, b]; the set of odd polynomials, 0, is not an algebra in C[a, b]. The polynomials with constant term is an algebra in C[a, b]. If [( <;;; Rn is compact, the set of polynomials in n real variables forms an algebra in C(J{).

°

A subset B <;;; C(S) separates the points of S if and only if for t, s E S, tics, 3f E B such that f(t) ic f(s). Example 2 'P separates the points of [a, b]; [; does not separate the points of [-1,1]; o separates the points of [a, b]. Lemma 3 Let £ be a vector subspace of C(S) that separates the points of's' and is such that 1 E £. Then given t, s E S, tics, and a, bE R 3f E £ su.ch that f(s) = a and f(t) = b. Proof: There is 9 E £ such that g(05) ic g(t). Put c = g(s) - g(t). Then the function f = (a - b)g + (bg(s) - aq(t)) . 1 E £ c

and satisfies f(s) = a, f(t) = b. Definition 4 A subscl £ of C (S) is railed a function space if £ is a vcrlor' spaCF and if f, 9 E £ implies f V 9 E £ and f II 9 E £.

and

Note that if £ is a function space, then whenever f E £, j+ If I also belong to £.

= fV 0, f- = (-1) V

°

Example 5 Let 'P £ be the collection of all piecewise linear, continuous functions on [0,1] (i.e., f E 'P£ if and only if f E C[0,1] and 3 a partition Xo < XI < ... < Xn of [0,1] such that f is linear on each subinterval [Xi_I,X;].) Then 'P£ is a function space but is not an algebra.

A..4. THE STONg WEIERSTRA.SS THEOREM

263

Lemma 6 Lei £ be a function space in thai contains the constant function 1 a.nd separates the points of S. Then given 9 E C(S) and to E Sand E > 0, 3f £ such that f(to) g(to) and f(t) > g(t) e \;It E S. Proof: By Lemma 3, \;It S 3ft E £ such that ft(to) = g(t o ) and ft(t) get). Since ft and 9 are continuous, there is a neighborhood lit of t such that ft(8) > g( s)- e \;18 E K Then : t E S} is an open cover of S and, therefore, there are iI" .. ,tn E S such that

Ulit,

S. Let f

i=1

= ftl

V··· V

Then f E £ and f(t o )

= g(t o ).

Also, if

E 8., then 3k such that 8 E lith' Then f( s) 2: ft. (s) > g( s) e; so f is the desired function. We can now give the lattice version of the Stone-vVeierstrass theorem.

Ii

Theorem 7 Let £ be a function space in C( S) that contains the constant function 1 and licpamtes the points of S. Then £ is dense in C(S). Proof: Let 9 E C(S) and > O. By Lemma 6, \;It E S E £ such that !t(i) get) and ft(s) > 9(s) - E \;Is E S. By the continuity of ft and g, 3 a neighborhood Ut of t such t.hat ft(s) < g(s) + E \;Is E Ut . Since S is compact,

3t 11 ••• , tn E S such that Since It. > 9 f(5) 'S ft. < g(5)

UUt. = S.

Put f

i=l

0,

f >

+ E.

g(s)-e

9 - c. Thus

f(8)

If s

g(8)+(,

ft1 II··· II ft n • Then f E £. S, then s E Utk for some k, so that. \;IsESor

IIf-glloo'SE.

In order to state the algebraic version of the Stone-vVeierstrass theorem, a lemma is needed.

Lemma 8 There is a sequence of polynomials {Pn} such that Pn (t) fort E 1].

->

Vi u,nifonnly

Proof: Set PI = 0 and Pn+1(t) = Pn(t) + ~(t Pn(t)2) for n 1. Clearly, each pn is a polynomial. t 'S 1. This certainly holds for n = 1; We first claim that: 0 'S Pn(t) 'S Vi, 0 assume that it holds for n k. Pk+l(t) 2: 0 and

Thus, the claim is established by induction. Since Pn(t)2 'S t \;It E [0,1]' it follows that {Pn(t)} i \;It E [0,1]. Put pet) = limpn(t). Since pet) 2: 0 and p(t)2 t, we have pet) = Vi, that Pn(t) -> Vi \;It [0,1]. The convergence is uniform on [0,1] by Dini's theorem ([DeS] 11.18). Of course, the conclusion of Lemma 8 follows directly [rom the \Veierstrass approximation theorem. We gave an independent proof in order to show that the vVeierstrass approximation theorem truly is a corollary of the Stone\Veierstrass theorem.

APPENDIX A.

264

Theorem 9 (Stone-Weierstrass) Let A be an algebra in C'(5) such that A contains the con8tanl Iunction 1 and separates the points ol S.

Then A is dense in

C(5). Proof; Theorem 7, it suffices to show that A is a function space. Let {Pn} be the polynomials in Lelllma We first claim that I A implies III E > O. Then gn = lin 0 (J2 I a 2 ) E A 8, let I E A be such that I f. 0, and put a Vn (Exprcise 1) and, since Pn(t) -, uniformly for 0 S t 1, g" --t jPla 2 Iflla

vi

Hence, If I = a(lflla) EA. But f V g = ~(f + g + If - gil, and f II g = a function space. in

II Ilco'

!U + g

if

.11); thus, A is

Corollary 10 (Weierstra8s Apl)roximation Thwf'cm) The polynomials

aTe

indeed

dense hI

C[a,bj.

A more general statement is given by Corollary 11.

Corollary 11 Let [{ C(K).

Rn be compact. The polynomials in 1H.'ariables aTe dense in

Finally, we should check the necessity of the various hypotheses in Theorpm 9. The algebra £ in C[ -I, 1] does not separate the points of , 1] and is not dellse in C[-I,n so this conditioll cannot be dropped. Because the algebra of polynomials in e[O.I] that vanish at 0 is not dense in 1], the condition that the algebra contains the constant function 1 cannot be dropped. As stated above I,he Stone Weierstrass Theorem is not valid for complex-valued {cit: 0 :::: I S 2,,} be the unit circle in continuous functions. Far an example, let T the complex plane and cOIIsider the fUIlction f(z) z a.nd allY complex polynomial

p(z)

=

f.

k CkZ .

Write

f [p] for f(e it ) [p(e't)]. Then

k=O

so Hence, f is distance at least 1 from any polynomial so the polynomials cannot be dense in the continuous complex-valued valued functions on T. [Those readers familiar with complex variables know that the uniform limit of any sequence of polyuomials OIl : Izi S l} mllst be analytic] ror the complex form of the Stone- Weierstrass Theorem, we need to add an additional condition, namely, that whenever a function f belongs to the algebra A then so does its conjugate f. [Note this property is missing in the example above.] We denote by Cc (S) the space of all continuous, complex-valued functions on S. We assume that Cc (S) is equipped with the sup-norm.

AA. TilE STONE-WElEH.'iIR.\.'i.'i 'I'IIJ-:OUElv! Theorem 12 (CompteJ: /«)J'W of Slol!I Wcil'rstrass). Lei A be an algebra ill ('c(.'-> I such that A contains the ("()TI"ftmi fu.lldioll 1, separates the points of S and i., S(lI)' that when f E A, 1 E A. Thcn A is dense in CdS'). Proof: Let AR be the of all real and imaginary parts of functions Wil,i 1, belong to A. Since nf = (f -I- ])/2 and If = (f 1)/2i, AR is an algebra ill ('(» which contains 1 and sepa,rates the points of S. Hence, AR is dense in Si,l< ,. A = {J + ig: J,g EAR}, A is dense in CdS). For further remarks on the Stone-Weierstrass theorem, see the article by M. SIOll'·. in Studies in Modern Analysis, Volume 1 in Mathematical Association of A 1/11 "'({ Studies in lv1alhematics, edited by R.C. Buck ([Bu]). Exercise 1. If A is an algebra in

show that A is an a1gebra.

Exercise 2. Is P a function space in C[a, bF Exercise 3. If [. is a function space in C{S), show that Exercise 4. Show that P [.

'£ is

a function space.

Example 5) is dense in C[O, 1].

Exercise 5. Let A be the vector space in U[O,l] generated by the functions 1. sin It, sin 2t,

.• ,.

[f E A if and only if f( t) =

Ok

sin kt for some

0i

E

R.] Show that

A is dense in C[O, Exercise 6. Show that the algebra generated by the functions {I, t 2 } is dense in C[O,l] but is not dense in C[~l, 1]. Exercise 7. Let S, T be compact Hausdorff spaces. If f E C(8), g E C(T), write f fg) 9 for the function (s, t) ---> f(.s )g( t). Show that the functions of the form L h gk (finite sum) are dense in C(S x 7'). Exercise 8. Give an example of a situation where Theorem 7 is applicable but Theorem 9 is not. Exercise 9. If Jk E BV[a, b] and that f E BV[o, bJ?

h

--->

J

uniformly on [a, b], is it necessarily true

Exercise 10. If 9 E e[O, 1] and (. > 0, show that 3ao, al, ... , ak E R such that

Ig(t)

ta]cJtl<(.,

"It E [0,

266

Exercise 11.

APPENDIX A.

Show that the polynomia.ls with ration,tl coefficients are dense

III

e[a,b]. Exercise 12. Let A be the set of all functions of the fornl Show tha.t A is dense in

bj.

Exercise 13. LE't A be the set of all functions of the form Ck

E R. Show that A is (blse in

qo,

Is A dense in C[ -1f 1

,n E N.

Ck

E R.

References [AS] P. Antosik and C. Swartz, Matrix Methods in Analysis, Springer-Verlag, Heidelberg, 1985. [Ap] T. Apostol, Mathematical Analysis, Addison-Wesley, Reading, 1975.

[As] J.M. Ash, Studies in Harmonic Analysis, Math. Assoc. Amer., 1976. [B1] S. Banach, Oeuvres, PWN, Warsaw, 1967. [B2] S. Banach, Oeuvres II, PWN, Warsaw, 1979. [Ba] R. Bartle, The Elements of Real Analysis, Wiley, N.Y., 1976. [Ba2] R. Bartle, An Extension of Egorov's Theorem, Amer. Math. Monthly, 87 (1980), 628-633. [Bi] G. Birkoff, Lattice Theory, Amer. Math. Soc., N.Y., 1940. [Boa] R.P. Boas, A Primer of Real Functions, Math. Assoc. AmeL, Providence, 1960. [Bo] S. Bochner, Additive set functions on groups, Ann. Math., 40 (1939), 769-799. [Bb] N. Bourbaki, Integration, Hermann, Paris, 1952. [Br] J. Brooks, The Lebesgue Decomposition Theorem for Measures, Amer. Math. Monthly, 78 (1971), 660-662. [Bu] R.C. Buck, Studies in Modern Analysis I, Math. Assoc. Amer., 1962. [Cal

c. Caratheodory, Measure and Integration, Chelsea, New York, 1963.

[Cat] F. Cater, Most Monotone Functions are not Singular, Amer. Math. Monthly 89 (1982), 466-469. [Ch] S.B Chae, Lebesgue Integration, Marcel Dekker, N.Y., 1980. [Cob] S. Cobzas, Hahn decompositions of finitely additive measures, Arch. Math.) 27 (1976), 620-621. [CL] E. Coddington a.nd N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hili, N.Y., 1955. [Co] D. Cohn, Measure Theory, Birkhiiuser, Boston, 1980. 267

REFERENCES

268

[DG] R. Darst and C. Goffman, A Borel Set which Contains no Rectangles, Amer. Math. Monthly, 77 (1970), 728-729. [DM] L. Debnath and P. Mikusinski, Introduction to Hilbert Spaces with Applications, Academic Press, N.Y., 1990. [DeS] J. DePree and C. Swartz, Introduction to Real Analysis, Wiley, N.Y., 1987. [DU] J. Diestel and J. Uhl, Vector Measures, Amer. Math. Soc., Providence, 1977. [Di] J. Dieudonne, Foundations of Modern Analysis, Academic Press, N.Y., 1960. [Do] R. Doss, The Hahn Decomposition Theorem, Proc. Amer. Math. Soc., 80 (1980), 377.

[Dr] L. Drewnowski, Equivalence of Brooks-Jewett, Vitali-Hahn-Saks and Nikodym Theorems, Bull. Acad. Polon. Sci., 20 (1972), 725-731. [D] L.E. Dubins, An Elementary Proof of Bochner's Finitely Additive Radon- Nikodym Theorem, Amer. Math. Monthly, 76 (1969), 520-523. [DS] N. Dunford and J. Schwartz, Linear Operators I, Wiley, N.Y. 1958. [Fe] M. Feldman, A Proof of LUBin's Theorem, Amer. Math. Monthly, 88 (1981), 191-192.

[Fre] G. Freilich, Increasing Continuous Singular Functions, Amer. Math. Monthly 1973 (80),918-919.

[Fr] R. French, The Banach-Tarski Theorem, Math. Intell., 10 (1988), 21-28. [Ga] G. Gaudry, Sets of Positive Product Measure in which every Rectangle is Null, Amer. Math. Monthly,81 (1974), 889-890. [Gi] D.P. Giesy, A Finite-Valued Finitely Additive Unbounded Measure, Amer. Math. Monthly, 77 (1970), 508-510. [Gr] J.D. Gray, The Shaping of the Riesz Representation Theorem, Arch. History Exact Sci., 31 (1984), 127-187. [Hah] H. Hahn, Set Funclions, University of New Mexico Press, Albuquerque, 1948. [Hal] P. Halmos, Measure Theory, Van Nostrand, Princeton, 1950. [Ha] T. Hawkins, Lebesgue's Theory of Integration, Univ. of Wisconsin Press, Madison, 1970.

[He] R. Henstock, Definitions of Riemann Type of Variational Integral, Proc. London Math. Soc., 11 (1962), 402-418.

REFERENCES

269

[HS] E. Hewitt and K. Stromberg, Real and Abstract Analysis, Springer-Verlag, N.Y., 1965.

[HY] E. Hewitt and K. Yosida, Finitely Additive Measures, Trans. Amer. klath. Soc., 72 (1952), 46-66.

[Ho] H. Hochstadt, Eduard HeIly, Father of the Hahn-Banach Theorem, Math. Intell. 2 (1979), 123-125.

[J] R.C. James, Reflexivity and the sup of linear functionals, Israel J. Math., 13 (1972), 289-300. [KO] S. Kakutani and J.C. Oxtoby, Construction of a non-separable invariant extension of the Lebesgue measure space, Annals of Math., 52 (1950), 580-590. [Kn] K. Knopp, Infinite Sequences and Series, Dover, N.Y., 1956. [KF] Kolmogorov and Fomin, Introductory Real Analysis, Dover, N. Y., 1970. [Ku] J. Kurzweil, Generalized Ordinary Differential Equations and Continuous Dependence on a Parameter, Czech. Math. J., 82 (1957), 418-,149. [LI] H. Lebesgue, Integrale, longueur, aire, Ann. Mat. (3), 7, 1902, 231-359. [L2] H. Lebesuge, Aleasure and the Integral, Holden-Day, San Francisco, 1966.

[Lew] J.W. Lewin, A Truly Elementary Approach to the Bounded Convergence Theorem, Amer. Math. Monthly 93 (1986), 395-397. [Lo] L. Loomis, An Introduction to Abstract Harmonic Analysis, Van Nost.rand, Princet.on, 1953.

[Mac] H.M. MacNeille, A Unified Theory of Integration, Proc. Natl. Acad. Sci. U.S.A., 27 (1941), 71-76.

[Me] R. McLeod, The Generalized Riemann Integral, Mat.h. Assoc. Amer., Providence, 1980.

[MI] J. Mikusinski, Sur une definition de l'integrale de Lebesgue, Bull.

Acad.

Polon. Sci., 12 (1964), 203-204.

[M2] J. Mikusinski, The Bochner Integral, Academic Press, N.Y., 1978. [Mu] M. Munroe, Introduction to Measure and Integration, Addison- Wesley, Reading, 1953.

[Na] l. Natanson, Theory of Functions of a Real Variable, Ungar, N.Y., 1955.

[NI] O. Nikodym, Sur les suites de fonctions parfaitement additives d'ensemble abstraits, C.R.A.S. 192 (1931), 727.

270

REFERENCES

[N2] O. Nikodym, Sur lea famiIles hornees de fonctiona parfaitement additives d'ensernhle ahstrait, Monalsch. fur Math. und Phys., 40 (1933), 427-432. [P] A. Peressini, Ordered Topological Vector Spaces, Harper and Row, N.Y., 1967.

[Pel I. Pesin, Classical and Modern Integration Theories, Academic Press, N.Y., 1970.

[Raj J. Randolph, Basic Real and Abstract Analysis, Academic Press, N.Y., 1968. [RR] K.P.S. Bhaskara Rao and M. Bhaskara Rao, Theory of Charges, Academic Press, N.Y., 1983.

[RN] F. Riesz and B. Nagy, Functional Analysis, Ungar, N.Y., 1955. [Ro] W.W. Rogosinski, Volume and Integral, Oliver and Boyd, London, 1952.

[R] J. Romero, When is £1'(p,) contained in U(p,)?, Amer. Malh. Monthly, 90 (1983), 203-206.

[Roy] H. Royden, Real Analysis, MacMillen, N.Y., 1988.

[Rl] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, N.Y., 1976. [R2] W. Rudin, Real and Complex Analysis, McGraw-Hill, N.Y., 1966. [Se] J. Schwartz, A note on the space L;, Proc. Amer. Math. Soc., 2 (1951), 270-275.

[Sir] W. Sierpinski, Sur les rapports entre l'existence des integrales, Fund. Math., 1 (1920),142-147.

lSi] G. Simmons, Introduction to Topology and Modern Analysis, McGraw-IIiIl, N.Y., 1963. [St] K. Stromherg, The Banach-Tarski Paradox, Amer. Math. l'yfonthly, 86 (1979), 151-161. [St2] K. Stromherg, An Introduction to Classical Real Analysis, Wadsworth, Belmont, 1981.

[Swl] C. Swartz, The Evolution of the Uniform Boundedness Principle, lyfath. Chron., 19 (1990), 1-18.

[Sw2] C. Swartz, An Introduction to Functional Analysis, Dekker, N.Y., 1992. [Sw3] C. Swartz, The Nikodym and Vitali-Hahn-Saks Theorems for Algebras, Amer. Math. Monthly, 82 (1975), 833-834.

[Ta] A. Taylor, General Theory of Functions and Integration, Ginn, Waltham, 1965.

REFERENCES

271

[TL] A. Taylor and D. Lay, Introduction to Functional Analysis, Wiley, N.Y., 1980. [TT] A. Tulcea and C. Tulcea, Topics in the Theory of Lifting, Springer-Verlag, N.Y., 1969.

[Vu] B. Vulich, Introduction to the Theory of Partially Ordered Spaces, WoltersNoordhoff, Groningen, 1967.

[Wa] P. Walker, On Lebesgue Integrable Derivatives, Amer. Math. Monthly, 84 (1977), 287-288.

[Wi] J.H. Williamson, Lebesgue Integration, Holt, Rinehart, and Winston, New York,1962. [Z] A. Zaanen, Linear Analysis, North Holland, Amsterdam, 1956. [Za] T. Zamfirescu, Most Monotone Functions are Singular, Amer. Math. Monthly 88 (1981), 47-49.

[Zy] A. Zygmund, Trigonometric Series, Cambridge University Press, 1968.

Index tI,96

$ [xj, 181

limEb 2

1111,103

Iimtkl 2 I' ..L iJ, 139 Ii, x iJ, 116 /1
j,152

£oo, 168 tk 133 dl"

u(S x 7), 116 supE,2 f,,, 49

r(x), 100

Is Idft , 82 limEk,2 limtk,2 I j1 I, 30 11 I, 3 I x 1,198 j1 ..L iJ, 34 j1 V iJ, 138 j1/\ iJ, 135 j1 1> 24 j1-a.e., H j1", 36 j1+, 28 j1-, 28

n"

-0

49 P(S),1

ba(A),223 ba,224 BI/la, b], 255 C:"'(Rn), 127

CdS),104 DiJ(:r),119

J2f(x),154 1 -< V, 228

1 * g,

125

t+,3

r,3

124

f(

IaI,96

-< 1,

228

L(X),174

IimEk, 2

limh,2 DI(x),154

DOO(j1), 218 LCO(E),218

11111',207 1111, 166

L°(J), 110 Ll(j1), 103 ~liI), 103

1111=,218

It", ,1

LP(j1),207 Ip(E),207 Il, 249 spt(f), 104

C,s, 2

k".,2 £1'(8), 209

hl, 2

e,209 t,167 inf E, 2

!U,2

X', 175

272

273

INDEX

1,

249 T,182 x(a),240 n ,l

e

N,1 Q,1 R,1 Rn,1 R+,1 Z,1 B(S),39

F(S), 197 Iz,237

.c{f}(s),128 M(It·),36 P.c,263

1n: f, 96 'Rz,237 S,250

S(2:), 193 SeA), 193 A r f(x),146 B(p, q), 123 B(S),167

ba(2: ; It), 223 ba(A), 135 BV[a, b], 199 BVo[a, b], 204 c,168 C(S), 167 Ck[O, 1], 199 Co, 169 Co(R),250 Coo, 169 CA , 1 C c (S),265 ca(2:), 226 d+ f(x), 1M d- f(x), 1M d+f(x),154 Lf(x),154 D n ,244 Ei,119 E., 119

fV g, 3

f

1\ g, 3

f(-,t),117 f(8,-),117 r(x),251

f.,

151

f",127

J

Jx, 191

L(J,1f'),96

L(X, Y), 174 LO(It),109 L I (It),167 Lh(It), 237 Lfoc(R"), 145

m,24 M.L,239 m n ,25

Mf(x),146 N BV[a, b], 233 rca(S),232 s,168 Sex, r), 145 T ' ,192 U(J,1f'),96 Var(J : [a, b]), 255 var(J: 1f'), 255 x- y, 235 x ..L y, 238 x V y, 197 x 1\ y, 197 XjM, 181 x+,198

x-, 198 X", 190 X~,

200

A a-Cantor 53 It-almost everywhere, 74 It-almost uniformly, 74 absolute value, 198 absolutely continuous, 130, 159 absolutely 166 adjoint, 192 Alexanderoff, 63

INDEX

274 algebra, 16, 263 anti-symmetry, 197 Antosik-Mikusinski,65 approximate identity, 126 arclength, 258 Arzela,97 Arzela-Ascoli, 261 average value, 146

B BCT,92 Bounded Convergence Theorem, 92 B-space, 166 Baire Category Theorcm, 259 Banach integral, 194 Banach lattice, 202 Banach limit, 195 Banach space, 166 Banach-Steinhaus, 179 Banach-Tarski Paradox, 11 Bartle, 117 below,2 Bessel's Inequalit.y, 240 Beta Function, 123 bidual, 190 Bochner, 137 Bohnenblust-Sobczyk, 187 Borel function, 73 Borel measure, 51, 61 Borel sets, 39 bound, 2 bounded, 2, 171 bounded above, 2 bounded variation, 255

characteristic functions, 1 Closed Graph Theorem, 183 closed intervals, 3 complement, 1 complete, 166 complete orthonormal set, 241 Condensation of Singularities, 246 converge, 166 converges, 2, 23 converges pointwisc, 3 converges to f in jHIlcasure, lOG converges uniformly, 3 convolution product, 125 countable additivity, 10 countably additive, 20 countably subadditive, 21, 36 Connting Measurc, 23

D

c·

Daniell, 89 DCT,87 decomposition property 200 decreasing, 2-3 Diagonal Theorem, 66 Diagonalization, 261 Dini derivates, 1M Dirac Measure, 23 Dirichlet kernel, 241 distribution function, 56 Dominated Convergcllcc Theorcm, 87 dot product, 23,5 double sequence, 5 Drewnowski's Lt'mma, 35 dual, 175 dual norm, 175

complete measure, 46 u-cornpact , 61 canonical map, 191 Cantor Set, 52 Cauchy in jt-measure, 107 Cauchy-Riemann integral, 97 Cauchy-Schwarz Inequality, 235 CGT, 183 change of variable, 161

p-essential sup, 218 p-essentially bounded functions, 218 Egoroff,75 equicontinllollS, 180, 261 equicontinllollS from above at 0, 211 equivalent, 169 Euler, 101

l

E

INDEX

F u-finite, 44 F. Riesz, 107-108, 200 Fatou, 87 Fejer kernel, 246 Fejer's Theorem, 247 finitely additive, 20 first category, 259 Fourier coefficients, 240 Fourier Inversion Theorem, 251 Fourier transform, 249 Frechet metric, 168 FTC, 98, 145, 149, 157 Fubini, 117, 153 function, 20 function space, 199, 263 Fundamental Theorem of Calculus, 98, 145, 157

G Gamma Function, 100 Gelfand integral, 221 greatest lower bound, 2

H Holder Inequality, 221 Holder's Inequality, 207 Hahn Decomposition, 33 Hahn-Banach Theorem, 186 Hardy-Littlewood maximal function, 146 Hellinger-Toeplitz, 248 Hewitt-Yosida Decomposition, 48 Hewitt-Yosida decomposition, 59 Hilbert space, 236

I {t-integrable, 81 {t-integral, 84 {t-integrable, 85 ideal, 202 improper Riemann integral, 97 increasing, 2-3 infimum, 2, 197 inner product, 235 inner product space, 235

275 inner regular, 51, 61 integral, 81-82, 96 integral operator, 217 intersection, 1 intervals, 2 inverse Fourier transform, 251 isometry, 175 iterated series, 6

J Jordan Decomposition, 29-30

K kernel, 217

L Laplace transform, 128 lattice homomorphism, 199 lattice isomorphism, 199 lattice semi-norm, 202 least upper bound, 2 Lebesgue, 152 Lebesgue Decomposition, 139, 161 Lebesgue integrable, 85 Lebesgue integral, 85 Lebesgue measure, 24-25, 4:3, 49 Lebesgue point, 147 Lebesgue set, 147 Lebesgue-Stieltjes, 24 Lebesgue-Stieltjes measure, 43, 56 Lebesgue-Stieltjes signed measure, 160 left shift, 176 Leibniz, 100 limit inferior, 2 limit superior, 2 linear functional, 175 linear topology, 165 linearly isometric, 175 Lipschitz condition, 159 locally integrable, 145 lower, 2, 96 lower variation, 28 Lusin, 78

M

INDEX

276 ;,t-negative, 33 ;,t-null, 33 ;,t-positive, 33 L:-measurable,71 ;,t x v-measurable, 116 Maximal Theorem, 146 MCT, 83 measurable, 71 measurable rectangle, 116 measure, 20 measure space, 20 metric linear space, 165 metric outer measure, 39 Mikusinski, 113 Minkowski Inequality, 208 Minkowski's Inequality for Integrals, 211 monotone, 2-3, 20, 36, 204 monotone class, 17 Monotone Class Lemma, 18 Monotone Convergence Theorem, 83 mutually singular, 31

N negative part, 28, 198 Nikodym Roundedness Theorem, 68 Nikodym Convergence Theorem, 67 NLS, 166 non-atomic, 222 norm, 165 normed space, 166 nowhere dense, 259

o OMT, 184 open intervals, 3 Open Mapping Theorem, 18·1 operator norm, 174 order bounded, 200 order completeness, 2 order dual, 200 order ideal, 202 ordered vector space, 197 ordinate set, 124 orthogonal, 238 orthogonal complement, 239

orthonormal, 238 orthonormal dimension, 2,13 outer measure, 36 outer regular, 51, 61

p pairwise disjoint, 1 Parallelogram Law, 2:~6 Parseval's Equality, 242 partial order, 197 partition of nnity, 228 permutation, 1 Pettis, 193 Pettis integral, 221 Plancherel transform, 251 Point Mass, 23 positive, 197, 199-200 positive part, 28, 198 power set, 1 premeasure, 20 product, 116 purely finitely additive, 48 Pythagorean Theorem, 2:39

R Radon-Nikodym, 131, 138 Radon-Nikodym derivative, 1:33 rapidly decreasing functiolls, 250 rearrangement convergent, 5 rectangle, 116 reflexive, 191,197 regular, .51, 61, 63 representing measure, 228 representing signed measure, 2:32 Reverse Holder Inequality, 212 Riemann integrable, 96 Riemann-Lebesgue Lemma, 249 Riesz,170 Riesz Decomposition, 200 Riesz Representation Theorem, 175, 213, 219, 224, 229, 23:3, 2:39 Riesz space, 197 Riesz-Fischer, 208, 243 right shift, 176 ring, 16

INDEX

S a-set, 15 a-algebra, 17 E-simple, 78 s-section, 119 t-section, 119 8-sequence, 126 scalar product, 235 Schwartz space, 250 second category, 259 second dual, 190 semi-algebra, 15 semi-norm, 166 semi-normed (normed) vector lattice, 202 semi-ring, 15 separates the points, 263 shrinks regularly, 148 sides, 116 Sierpinski, 121 signed measure, 20 simple, 78 singular, 34, 139, 153 standard representation, 78 Stone-Weierstrass, 265 sublinear functional, 186 subseries convergent, 7 sum, 96 sup-norm, 167-168 supremum, 2, 197

T Tonelli, 119 topological vector space, 165 total variation, 30, 256 transitive, 197 translation invariant, 51 transpose, 192

U UBP, 178 unconditionally convergent, 5 Uniform Bounded Principle, 178 uniformly II-continuous, 134, 142 uniformly II-continuous, 211 uniformly count ably additive, 66

union, 1 upper, 2,96 upper variation, 28

V variation, 30, 255 vector lattice, 197 vector sublattice, 202 vector topology, 165 Vitali,211 Vitali-Hahn-Saks ThcorclI1, II'!

w Weierstrass Approximatioll '1'1.,,<., <'"

'",